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HENRY HOM AN D COMPANY 


NEW YORK CHICAGO 


SCHOOL ALGEBRA 


SECOND COURSE 


BY 
is hod Rd od 8 ed BVA aa ered DB 


UNIVERSITY OF ILLINOIS 


mee CRA THORNE, Pu. D. 
UNIVERSITY OF ILLINOIS 
AND 


Hue, LAY LOR, ‘PH.D; 


EASTERN ILLINOIS STATE NORMAL SCHOOL 





NEW YORK 
HENRY HOLT AND COMPANY 
1915 





* 
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“wr 
at 4 
E : 
, ‘ 
as 
ret - 
ae 





‘COPYRIGHT, 1915 
| a 
HENRY HOLT AND COMPAN 





uy se 
1919 MATHEMATICS LIBRARY 


PREFACE 


This Algebra is the second volume of a series of two books. 
It contains material for a half-year course and presupposes a 
year’s work such as is provided in the authors’ First Course, 
or any similar text. Since, in many schools, a year or more 
intervenes between the first and second courses in algebra, it 
has been deemed advisable to include a review of the First 
Course before proceeding to new subjects. This part of the 
book is, however, more than a mere hasty resumé. While the 
student is reviewing the topics of the first year, he is at the 
same time making a distinct advance by seeing the subject 
from new. view-points, which his added maturity and training 
enable him to appreciate. 

The selection of new topics in this Second Course is in accord 
with current practice in the best high schools. While the book 
contains material for thorough preparation for college, the 
authors have not had in mind solely the students making such 
preparation, but rather the great body of students who will go 
no further in our educational system than the high school. 
The chapter devoted to the notion of a function is introduced in 
the belief that the high school student should become somewhat 
familiar with the idea of the correspondence between two re- 
lated variables. Along with the notion of a function, the graph 
is presented with a view to leading the students to picture 
and to visualize this correspondence. 

A special effort has been made to obtain interesting problems 
that have an inherent value even when considered apart from 
the algebraic principles involved in their solution. Without 
aiming to teach physics or engineering, some of the problems 


lv . PREFACE 


have a physical setting; but, in such cases, care has been taken 
to assume no more technical knowledge than is possessed by 
the average high school student. A feature of the numerous 
review exercises and problems is the inclusion of some 
typical college entrance examination questions. 

The authors wish to express their thanks to the teachers who 
have contributed. to this book by their suggestions. We are 
especially indebted to Prof. E. J. Townsend and Dr. E. B. Lytle 
of the University of Illinois and to Miss Jessie D. Brakensiek 
of Quincy, Ill., High School, for careful and critical reading of 
the manuscript and for suggestions as to exercises and problems; 
to Mr. J. L. Dunn of Lewis and Clark High School, Spokane, 
Wash., Mr. C. H. Fullerton and Mr. W. B. Skimming of East 
High School, Columbus, O., Mr. E. A. Hook of the Commer- 
cial High School, Brooklyn, N. Y., Mr. L. C. Irwin. of the 
Township High School, Joliet, Ill., and Mr. R. L. Modesitt 
of the Eastern Illinois State Normal School, Charleston, IIl., 
for reading the proof and seeing the book through the press. 


* i. eRe 
A. R. CRATHORNE 
E. H. TAYLOR 


CONTENTS 


CHAPTER I 
ALGEBRAIC OPERATIONS 


ARTICLE PAGE 
ier our vondamental Operations. . . 2.0. . ste ee ew 1 
ret Ubtraction . . . . ww woe 8 we lee we 2 
MUPIIPROTUreNGHeSeS:. . . kk kk i kh ee we 3 
4, Multiplication. .... Poe is codes Labnre as | Se 5 

TTR ee we ee 6 
6. Multiplication of Rate wninls See ish. kc ease Pe cast, es 6 
7. Multiplication of Polynomials by Noemi SS; eR ea <s 6 
8. Multiplication of Polynomials . fi 
9. Zero 7 

10. Division : 8 

11. Division of Peremials : ow ; are 

12. Division of a Polynomial by a Mecamial ‘ 9 

13. Division of a Polynomial by a Polynomial 9 

SRI MIVISION 1 ek i 9 

PTT UO OS. kk kt we thie te 11 

MeMerICMOPeVAtlONs .; =< . i «>. cous ws te et te 12 

CHAPTER II 
LINEAR EQUATIONS IN ONE UNKNOWN 

Eg SS et Pe er ak eer ee 15 

18. Equations and Petes as Pentcnces > 1, ideale eee ae a 

Peon or Hoot of an Equation. ...... 4... . eo ee 

MenGmetivmiraieng quatiOnNS . . . . . 1 1 ee tk ww we 16 

Derren OT TAQUALIONS .~. . . . - «ss 6 8s 0 eee 1; 

22. Linear Equations in One Taino . ew tye 18 


23. Verification by Substitution... 2... 6 ee ee ee 18 


vl CONTENTS 


CHAPTER III 


FACTORING 
ARTICLE PAGE 
24. Rational Integral Expressions .. . 5 « « . si\smin: sen 25 
26. Definition of Factoring . . .°. . «2s «© 0s sneene 25 
26. Important Special Products . .) ... . 4. » % en eee 26 
41.2 be urthersey pe Products 20.5 oo a) sna eee Pe 29 
28, Factor Theorem . >. . . 4%. Su) 2) so 30 
29. The Sum or Difference of the Same Two Powers. ...... 32 
30. Summary of Factoring’ .-°.) 2) 2 34 
31. Solution of Equations by Factoriige oes ce ene 38 
32. Highest Common Factor . . ...°5. 5. 2) seve 38 
33. Lowest Common Multiple. . 1... ... . 1: sts eee 40 

CHAPTER IV 

FRACTIONS 
$4. Algebraic Fraction’. ... .o 0a 3. %. « cats 6s 4} 
36. Reduction of Fractions . . . .-.. 55>. . . eee 41 
So, heduction to Lowest Terms:< <=). 05-5 6s ee Ye) 3 42 
37; Addition and Subtraction. . . 4%. 5... 2°.°. =.) nn 44 
38. Multiplication and Division . . . . . « . . 94 4 46 
39. Complex Fractions... . ak sa) a 48 
40. Equations Involving F pcre ame 50 

CHAPTER V 

FUNCTIONS 
41. Constants and Variables .. . ... . 2. . . ee eure 
42. Definition of a Function ....... < « . «nn 59 
43. Functional Notation . 2... 9... .. . . . . ne 60 
44. Evaluation of Functions 61 
45. Formulas Frequently Used “5... . =. . . 13 62 
46. System of Coérdinates JF... . . 2.) (on 66 


47. Graph ofa Function . ee... dS 68 


CONTENTS 


CHAPTER VI 
SYSTEMS OF LINEAR EQUATIONS 


ARTICLE 


48. 
. Graphical Solution of a System of Linear Equations .... . 


mmeneranear Mquations ~. .-« +s. «1 + et be ft 


NISSIM Fs Ch yg - ey vem deen ee ge elk cde Ry 


. Elimination by Addition and Subtraction. ......... 


Pemesrurnstnstivution 6. ek A wows 
System of Linear Equations Solved by Determinants... . . 
Perera Os CNG.) bird Order «<< gel ee 


. Solution of Three Equations with Three Unknowns ..... 
MeeerOIenG EATUAGIONS . 2 4 6 ky te te we 


CHAPTER VII 
RATIO, PROPORTION AND VARIATION 


RES a ea 
TeLarieumVvolved.in Wieasurement . .. . . . 6 65 6 a 8 
PMU et ee ee es 

MM Te een TOOOTIIONS 4... se ke 
ee ee 
RE ee sk a aa we a es 
SS RCO a a 


CHAPTER VIII 
EXTENSION OF THE NUMBER CONCEPT 


eM eUTaCtIONSs Joo 6 8). ew ee ee 
MEERREMMMTUOOTS © § (05: ys, 5) spe ee (sb wen op te Slows 
ITE tiers FF ta eo es ee 
Ieee gg Py SARTO oe Se te i OG 
sO” PS RP Rr ea aa 


SEMeetee sche lL tg Oh ae alg es BS 


. Products of Imaginaries. ... . 53 ew SE a 
DONTE Sk ew ee 
. Operations with Complex inne on ee BS Se 
meenmiueate Compiex Numbers .....: .. +... + Gee 
. Graph of a Pure Imaginary Number . . .. .’< te 
. Graph of a Complex Number .....--. +++ e+e ees 


vil 


Vill 


CONTENTS 


CHAPTER IX 
QUADRATIC EQUATIONS 


ARTICLE F . : PAGE 


77. Typical'Form . . 1°... 6 « « + # ts) nots 123 
78. Solution by Factoring: . ... . . si. 5 + > eee 124 
TomeeoulLion py. Hormula, .. . a 8 ee 0 a 125 
80. Special or Incomplete Gusdeates reer? ? roe 128 
81. Formation of Equations with Given ear . 9 ee Aer ase 
82. Nature of the Roots of a Quadratic . . . .. 5 Vj 
83. Graph of the Quadratic Function .. 5 . 7g 134 
CHAPTER X 


84, 
85. 
86. 
87. 
88. 


89. 
90. 
91. 
92. 
93. 
94. 
95. 
96. 
97. 
98. 
99. 
100. 
101. 
102. 


SYSTEMS OF EQUATIONS INVOLVING QUADRATICS 


Introduction =e ee 141 
One Equation Linear el Gna Quadtratie - 2 ee +e ee 
Both Equations Quadratic .. . oe + a 145 
Both Equations of the Form az? + by? ae c=0. . 2 2 ae 145 
Further Special Methods .... .% . . 5. en 146 
CHAPTER XI 
EXPONENTS, RADICALS AND ROOTS 
Positive Integral Exponents ... . . + ey 2 
Extension of the Meaning of an Ranonert: <a ~ Kay oe 
Meaning of a@® . ... . 2 - 6 © +e 6 eee 154 
Meaning of a PENERES PR Poow st Mot 
Meaning of as as er 
Meaning of am wheat m is neeauee we eee 
Rational and Irrational Numbers . ......4.%4 «+ seo 
Radicals 2.) 6 ee ue ee ee ee 
Surds a: aw . + 02 |) ol igen et kee 
Introducing a Bor Bist: ane Festina Sieh Me ee 
Simplification of Radicals . 32)... - ». . |... ene 
Simplest Form of a Radical .. . ~ + 0 + Teepe es 
Addition and Subtraction of Radic: AN eee . » bee 
Radicals Reduced to the Same Order ........, . & eee 


CONTENTS 1x 


ARTICLE PAGE 


Samer mictoicanion Of Radicals ... . 2.4. + . 2 «6 2 6 « » eve 166 
IEEE ECAUICAIS Oy gk eS wt ae ee is? ee we 169 
muememcuare toots of Polynomials . . 2°. 5... 2. we ee 172 
106. Square Roots of Numbers Expressed in Arabic Symbols . . . 173 
107. Square Roots of Radical Expressions of the Form a+ /6. . 173 
ree ttO0ts OF Polynomials <2... 2 2. KR ke Sa 174 
109. Cube Roots of Numbers Teeeseers in Arabic Symbols . ... 176 


110. 
rh 
112. 
113. 
114. 
115. 
116. 
117. 
118. 
119. 
120. 
121. 


122. 


123. 


124. 
125. 


CHAPTER XII 


LOGARITHMS 

Seeerenigien Doraritnm. . °°. . ow Ss dlygeeed Merc ees 179 
Numbers Expressed as Powers of Given Numbers .... . 179 
moerectenisie and Mantissa . ..°. 6 5 6 shou. Whe. GS 180 
IIIT @LCHR OE POCUCH: cea voeeal se Se Lee Bale 181 
Peer inerurs COUIeNb. 20. oS ee eo ee i8l 
MPU CR IE OWEE = cee 8 ee ke wh ke we 181 
REESE IGUTNE ooh) el Se Gk ny We le ek) wae 183 
Determination of the Gharaceerintic oF a ence Logarithm . 184 
Mantissa of a Common Logarithm :. ........ we carat” aieses 
Prsonory=loggr (a>l). ... . 86 
To Find from the Table the Copentnn ‘of a ara Nanna a Tre 
To Find from the Table the Number that Corresponds to a 

PEPE LERSIITYS She Sar, SMa Var ics sha ts ono dete ss . 189 
Computation by Means of Logarithms .......... 190 


CHAPTER XIII 
EQUATIONS INVOLVING RADICALS 


Solution of Equations Containing Radicals. ........ 200 


CHAPTER XIV 
EQUATIONS IN QUADRATIC FORM 


i Si err irers ci ceo 204 
pameemeeanieous Quadratic Form .......++.+s+«see 207 


CONTENTS 


CHAPTER XV — 


PROGRESSIONS 


ARTICLE 


126. 
127. 
128. 
129. 
130. 
131. 
132. 
133. 
134. 
135. 
136. 
137. 


138. 
139. 
140. 
141. 
142. 


Arithmetical Progressions ...... 


ee ee ee ee ee he 


Elements of an Arithmetical Progression . . 


Relations among the Elements ... . 
Arithmetical Means 2-2) a. ee oe 
Geometrical Progressions .. . sige 
Elements of a Geometrical Brieresnicn 

Relations among the Elements ... . 
Geometrical Means: .. 4.48%, seas oun 
Number of Terms Infinite ...... 
Repeating Decimals. ........ 
Harmonical Progressions ...... . 
Harmonical’Means™. =") 72 = 4) cae 


CHAPTER XVI 


© 8 @€° 0° Se) Gy eee see 


* «© 6° F £8 “Sel? & 6 eee 


e 8 © (6 #8 6.58) BAe 


e 8° 688 (Ste; ete) ae eee 


eo “ease & (SF ee Te) eee 


ee, ee a 


@-  @ «= 6! Se 1S) at See eee 


THE BINOMIAL THEOREM 


Factorial... Wee coi 2a 
Powers of -a: Binomial). 330-e aeeee 
The Expansion of (a+b)” 


. ¢€ sos © @& Set Oe ae eee 


Se a 


Formula for the rth Term of the Henan as (a +b)” 


Binomial Formula, Any Exponent . . 


INDEX 


PAGE 
210 
210 
210 
211 
212 
213 
213 
213 
215 
217 


217 © 


218 


221 
221 
221 
224 
225 


231 


SCHOOL ALGEBRA 
SECOND COURSE 


A LIST OF SIGNS AND SYMBOLS 


+, read plus. —, read minus. 
X, or -, read times. 

+, read divided by. 

=, read is equal to. 

=, read is identical with. 

~, read is not equal to. 

<, read zs less than. 

>, read is greater than. 

S, read is less than or equal to. 


=, read is greater than or equal to. 





a! or |a, read factorial a. 


( ) Parentheses. 
[ ] Brackets. 


{ } Braces. 


Signs of aggregation. These signs are used to col- 
lect together symbols which are to be treated in 


operations as one symbol. 





Vinculum. 

dy, read a subscript r, or a sub r. 

x’, x’ ..., read x prime, x second, ... respectively. 
a: b, read a ts to b. 

lim x, read limit of x. 

x=o, read x becomes infinite. 

logan, read logarithm of n to the base a. 

a”, read a to the nth power, or a exponent n. 

a, read square root of a. 

\/a, read nth root of a. 

f (x), F (a), ete., read “f£’ function of x, “F” function of x. 


(x, y), or P (a, y) read point whose coordinates are x and y. 


SCHOOL ALGEBRA 


SECOND COURSE 


CHAPTER I 


ALGEBRAIC OPERATIONS 


1. Four fundamental operations. In the first course in 
algebra, it has been seen that operations with numbers are 
made up of additions, subtractions, multiplications and divi- 
sions. These operations are called the four fundamental 
operations of algebra. 


EXERCISES 
1. If a = 2, name the fundamental operations used in find- 
4 4 
ing the value of a? + 2a +1; of a — fe + 2. 


2. What fundamental operations are used in finding the 
aoc ot — f 





value of when x = 2? 
x+2 
3. The area of a trapezoid is $(b + )’)h, where b and Db’ 
are the bases and h is the alti- 1! 


tude. What operations are 
used in finding the area when 
be = 10, h =“8? 

4. The safe load W that 
may be put on a pile so that 
‘the pile will not sink under it 
is given by the formula, 





2 ALGEBRAIC OPERATIONS [Cuap., I. 


where W is the safe load in pounds, w is the weight of the pile 
driver hammer, h the distance in feet the hammer falls, and k 
the number of inches the pile goes in at each stroke of the 
hammer. What fundamental operations are involved in this 
formula? 


2. Addition and subtraction. The sum of two or more num- 
bers is the same in whatever order they are added. Thus, 


a+b=b6+4a4. 


This is the law * of order or the commutative law, for addition. 
The sum of two or more numbers is the same in whatever man- 
ner the numbers are grouped. ‘Thus, 


a+b+c=(a+b)+c=a+(b+0o). 
This is the law of grouping or the associative law, for addition. 


Example 1. Add 5a + 22, 4x — 6a, 8a — x. 
The sum may be written thus: 
(5a + 22) + (4% — 6a) + (Sa —2). ee 
By the associative and commutative laws, we may write (1) in the form 
5a +8a —6a+20+4n—2. (2) 
Again, by the associative law, (2) may be written as 


1a + 5a. 


This example illustrates the fact that polynomials may be 
added by combining similar terms. For this purpose, it is 
well to write the polynomials so that similar terms are in the 
same column. 


This process applied to Example 1, gives: 
5a + 2% 
—6a + Ax 
8a —- 2 
7a + 5a 


* These laws of algebra are not to be proved. They are in the nature 
of useful assumptions that have a wide application in experience. 


Anns. 2, 3] USE OF PARENTHESES 3 


. Definition of subtraction. To subtract a number a from a num- 
ver b is to find a number which added to a will give the sum b. 


Example 2. Subtract 522 —6x + 4ax from 8x2 — 4ax + 2ze. 
PROCESS: 8x? — 4ax + 2x 

oz? + dar — 62 

3x? — 8axr + 8x 





3. Use of parentheses. In order to group terms together, 
wwe use parentheses. It should be remembered that parentheses 
may be removed with or without change of sign of each term 
neluded, according as the sign — or + precedes the parentheses. 


Thus, a—(b-c)=a-b+4¢e, 
' a+(b—c)=a+b-e. 

In case several parentheses or signs of aggregation are in- 
rolved together, one within another, they may be removed 


dy removing the innermost ones in turn, according to the rule 
‘ust stated. Thus, 


a — {3x — A[x — (y — 5) — (2y + 6)]} 
=a — {3x —4[x —y + 5 — 2y — 6]} 

— {3u — 4a + 12y + 4} 
a+ax— 12y —- 4. 


Add: EXERCISES 
1. 4axz, —2az, bax. 
2. 4x + dy, 2x — by. 
3. 5cx + 6a, 4cx — 8a. 
4. 2x + 3y — 62, 3a — 8y +. 72, 8x — Ty + 4a. 
5. 4a? + d5ac” — 3c’, 8a? — Vac”, 2ac? + 5c’. 
6. Qa? — 382xy, 16x? + 172ry. 
jai & + 40x + c; 
| — 3x? — Zax + 3c, 


Tae — 2c. 





4 ALGEBRAIC OPERATIONS [Cuap. I. - 


Subtract the first expression from the second in the following : 
8. 3a +b, 4a — b. 
9. 2x? + 3a — 4, 4a? —-4 + 5. 

10. 32? + 4ry — by? — 6a, 5x? — zy — y? + 62. 

11. 4a? — Gar? + 2ax — 3a + 4, 2a? + ax — 8ax* + a —6. 

12. 3(a — b), 5(a — b). 

13. Show how you can check a subtraction by means of an 
addition. 

14. From the sum of 6a — 7b + 4c and 8b — 4a + 2c, sub- 
tract the sum of 2a — 5b + 2c and 6a — 3c + 5b. 

15. From the sum of az? + 3y + 42 and 3ax? — 8y + 10z, 
subtract 2ax? + 7y + 102. 

16. From the sum of 8(a—b) + 5(c—d) and 10(a — b) 
— 6(c — d), subtract 4(a — b) — 10(c — d). 

Combine coefficients of similar terms: 

17. aw + bx 4+ ce. 

SOLUTION : ax + bx + cx = (2 +6 + 0)2. 

18. 2ax — 6bx + 4cz. 

19. 4c — abx + cx. 

20. m(a+ 6) +n(a+b). 

SoLuTION: m(a+b) + n(a+ b) = (m+n) (a+ BD). 

21. m(a+b—c)+n(a+b-—c)4+l(a+b-c). 

22. 3a(5m — 3n) + 5a(—3n + 5m). 


Simplify by removing parentheses and combining like terms: 
23. 4a —5 + (y — 2a) + 3(y — Q). 

24. 8« + (82 — 5) — (67 — 11). 

25. 6y — 2(y — 4) + 3(y — 5). 

26. 2(7 — 5) + 6(x + 1) — 8(@@ — 7). 

27. 5x — 2[2 + 3(a — 1) + 5(@ — 2) — 4(a — 5)].* 


* For names of various symbols of aggregation, or of other symbols, ) 
see page facing Art. 1. 


Arts. 3, 4] MULTIPLICATION D 


Somer tee) — {27 — yy — (2 yy — 2) + Dr) — By]. 

29. a — (-2b —c — {a — b}) — (5a —.2c — [—3b + c]). 

BOs — (—2 — { —32:— [+ — x + 3] — 4} — [22 — x — 3)). 

31. 6% — 2[2 — 3(2x — 3 — a) — 5{a — (Qa — 2a) — 4}]. 

32. 7a — [(3a — 4) — 6 — (8a — 4c — 2)]. 

33. 5x — 4[2(a — 4) + 3(22 — 1) —- (a — 7). 

Simplify the following by combining like terms: 

Bae 1) 4 3 4/x? — 1. 

35. DV a? — a? + 2/2? — a? ~— 3V/2? — a’. 

36. mv/a? — a? + nv 2? — a?. 

eee tin — 2) + (v7 +1) (n — 1) 

38. n(n — 1)(n4+ 2) + n(n — 1)(n+ 1) + n(n — 1). 

In finding the value of this expression for n = 3, is it simpler 
to evaluate the three terms separately or to combine them 
first? 


39. State the rule for enclosing terms in parentheses pre- 
ceded by a minus sign. 


4. Multiplication. The factors of a product may be taken in 


any order. Thus, ab = ba. (1) 
_ This is the law of order or the commutative law, for mul- 
tiplication. 

The factors of a product may be grouped in any manner. 
Thus, abc = (ab)c = a(be). (2) 


This is the law of grouping or the associative law, for 
multiplication. 


To illustrate the use of the laws of order and grouping, we may write 
(5ab) (Zab) = (5- a+ b)(3-a-b) 
=5-3(a-a)(b- b) 
= 15a7b?. 


6 ALGEBRAIC OPERATIONS [Cuap. I. 
5. Exponents. We write 
a-a.... (to m times) = a”, 


and read a” as “‘ a exponent m”’ or the “ mth power of a.” 
When m and n are positive integers, 


a”™- a" =a-:a...(mtimes)-a-a... (nm times) 
=a-:a-:a-:...(m+n times) 
ee, 
or a"a® = art, (3) 


That is, the exponent of the product of two powers of a number 
is the sum of the exponents of the powers taken singly. This is 
often called the first law of exponents. 


ILLUSTRATIONS: a? - a® = a’, 3?- 34 = 38 5- 5?- 5 = 5%, 


6. Multiplication of monomials. The main points involved 
in the multiplication of monomials are well illustrated by a 
simple example. Thus, 


(6a*b*) (—8a?b) (4ab) 
(6- —3- 4)(a?- a? - a)(b?-b- 6b) [(1) and (2), Art. 4.] 
—72a*b+. 

It is to be noted (1) that the coefficient in the product is the 
product of the coefficients in the given factors, (2) that the 


exponent of any letter in the product is the sum of the expo- 
nents of that letter in the given factors. 


I 


7. Multiplication of polynomials by monomials. It is a 
law of multiplication that 


abb+c+d+...)=ab+ac+ad+... (1) 


This is known as the distributive law. 

According to this law, to multiply a polynomial by a monomial, 
multiply each term of the polynomial by Ue monomial and add the 
resulting products. 


Arts. 8, 9] ZERO ve 


Thus, 3a7b2 — 2ab? + 3a%b 
Bab 
9a%b? — 6a2b? + 9a3b2. 





8. Multiplication of polynomials. To find the product of 
two polynomials, we take the sum of the products obtained by 
multiplying each term of one polynomial by each term of the other. 

To illustrate, multiply 2a? + zy + 4y? +y by x —y + 2y. 

PROCESS: Qu? +xzy +4y? +y 
co —-y +2y 
203 + xy + 4ry? + xy 
. — 2x*y — xy? — 473 -— y? 
xy" + 2xy + xy? + 4ary 
223 — xy +4ry? + ry — 4y? — y? + 2a3y + wy? 4+ 42ry’. 





This process is, in the main, an application of the distributive 
law of multiplication. 


9. Zero. ‘The number zero may be defined by 
0=a-a, 
where a is any number. 
Then, it follows at once that the product of any number k 
and 01s 0. For, 
k-0 = ka — ka (By the distributive law) 
= (); (By the definition of zero) 


EXERCISES 


Perform the multiplications indicated in the following : 
Pie © 2°); 0°. 

2ab? - 3a*b? - 4a°*b. 

Ary - 30°y - — 2axy?. 

3abe - — 4ab’c - 2abc’. 

On A gl 4 ghd, 

gn . gn tl : gr —t, 

Drn+tm+1 F 3r—m—1 5 Dm—n—1 ' gm tn— 1 


See aye (1 +x)" 1 ae 


ek Ee GMAT soa ala a 


8 ALGEBRAIC OPERATIONS (Cuap. I. 


9. 2a(4a? — 3a). 
10. 5a?(4ab + 3a7b — 2a%b?). | 
11. (2a? + 3ay) (3a? — 2ay). 
12. (a+ b)(a+ b)(a+ 0). 
13. (1 —x)(a + ax + az? + az’). 
14. (a+6+4¢)?. 
15. (a+b — Cc)’. 
16. (a+b+c)(a+b-c). 


b b 
3 
18. (22 - u) 


10. Division. T'o divide any number a by any number b 
(6 4 0) * is to find a number x such that the product ba = a, 

Note the condition that 6 # 0. This means that the divisor 
is not zero. | 

From this definition of division, 


a a’ . 
= da, at a’, 
a a 
¥ a™ 
and in general — = ne, (1) 


a” 


where m and n are positive integers and m > n. 
That is, the exponent in. the quotient of two powers of a number 
is the exponent of the dividend minus the exponent of the divisor. 


11. Division of monomials. The main points involved in 
the division of monomials may be brought out by a simple 
example. 


ae 4f})2 4 2 
Fre. 6a‘b os; a b 





Sa eee) ab 
= —2a*b. 
It is to be noted (1) that the coefficient of the quotient is the 
quotient of the coefficients with due regard to the rule of signs, 


* The sign ~ is the sign of inequality and is read “is not equal to.” 


Arts. 11, 12, 13,14] ZERO IN DIVISION 9 


(2) that the exponent of any letter in the quotient is equal to 
the exponent of that letter in the dividend minus its exponent 
in the divisor. 


12. Division of a polynomial by a monomial. According 
to the distributive law (Art. 6), a polynomial is divided by a 
monomial by dividing each term separately. 


622y — dary + Yary? 





Thus, Bry = 2x —a + day. 
Cueck: Makex =y=a=1. Dividend = 12. 
Divisor = 3. 
Quotient = 4. 
But 12 dl 


13. Division of a polynomial by a _ polynomial. The 
process of dividing one polynomial by another may be easily 
shown by examples. In general, the dividend and divisor 
should be arranged according to ascending or descending powers 
of some letter, called the letter of arrangement. 


Example. Divide a‘ — 7a? + 2a* — 8a + 12 by 2 + a? — 3a. 


PROCESS: a4+2a3 — 7a2-— 84+12 | a? —3a4+2 
at — 3a° + 2a? a? + 5a+6 
5a*— “9a? — 8a+4+12 
5a? — 15a? + 10a 
6a? — 18a + 12 
6a? — 18a + 12 








Check this division by finding the product (a? + 5a + 6) (a? — 3a + 2). 


14. Zero in division. In the definition of division (Art. 10), 
it was stated that the divisor cannot be zero. That is to say, 
division by zero is excluded from our operations in algebra. When 
the dividend is zero, the quotient is zero ; for, 

0 
ay 


since k-0=0. (Art. 9) 


10 ALGEBRAIC OPERATIONS [Cuap. I. 


Where is the fallacy in the following ? 


Let t= 0. (1) 
Multiply both sides by z, x? = az.: (2) 
Subtract a? from both sides, x? — a? = ar — a2. (3) 
Divide both sides by x — a, x+a =a. (4) 
But by (1) 2 =a. (5) 
By (4) and (5) 2a =a. (6) 
Hence, 2=1. (7) 

Divides EXERCISES 

1.23 = 31. 7oihyD.. Bye | 

2. 3a’b’c by abe. 

Dal eye. 

4. qe + 3d + 2 by qe +dadt 2. 

5. a" — 5y2n +1 by gnr— syn —1, 

6. (x — 1)*(@ + 1)"*! by (@ — 1)*"(@ + 1)". 
7 (& — y)(a + y)” by ( — y)®(a + y)>. 

8. (a2 — b2)2"+38 by (a2 — b2)2, 

9. xr y* — 5x2yt — xy? by —x?y?. 

10. vs? — 2gsv — 2s?v by vs. 

11. St? — 3g by ig. 

12. 12a7¢? + 9xtt® — 1523 by 3x02. 

13. 9Yax*w — l5ax?w? — 8artw* by 3ar2w. 


a 
iw 


. Sap*g’ + 24bp?¢? — 32p%q? by 8p2q?. 

. —ldx?yz — dry’z — 1l5ryz? by 5zryz. 

16. a? + 3a*b + 3ab? + b8 by a? + 2ab + b?. 
Check your result by making a= 1, b= 1. 


- 
ao 


17. a — ab — ab? + b8 by a? — 2ab + B?. 
Check your result by making a= 2,b=1. Why ean this 
result not be checked by making a = 1, b = 1? 


18. 2 —-y by x-y. 
19. z+ y° by a+ y’. 
20. 4y?+ 4xy + 2? — 12yz— 6x2 + 92? by 2y + x — 3z. 


Arts. 14, 15] POWERS AND ROOTS 1] 


Check Exercise 20 by making x = 1, y = 2, z= 1. 

21. 24+ 32° + 7274+ 824+ 6 by 224248. 

22. (a—b)?— 3(a—b64+2 bya—b-1. 

23. z*— yt by r—- y. 

Check the result by making x= 2,y=1. Why can the 
result not be checked by making x = y = 1? 


24. a‘ + 4ab® — 4abe + 3b* + 2b?c — c? by a? — 2ab + 3b? — . 
25. A man owns three farms whose values are 
oat + 5a + 277 +2 +5, 
5a* + 2x3 + Ox? + 8x — 7, 
and Aart + 823 + Tx? + 2x + 2. 


Find the value of the three together. He divides the land into 
parts of equal value among his x sons. What is the value of 
the land that each obtains? | 

26. Divide 7R? — mr? by TR + Tr. 


15. Powers and roots. In exercises under Art. 4—11, we 
have performed operations with integral powers by means of 
the laws of exponents 


a™a" =amtn. (1) 
and | a” > 
gh as (2) 


We shall soon find it convenient to use a third law of exponents 
(ann = ann (3) 


To illustrate the law, (3*)? = 34 - 34 = 38. 
To derive this law, we note that 


eye — aja" a” > ...to n times 
= qrntm+m+-->+-+ton terms 
=a, 


The pupil knows something about square root and cube 
root. We may now well extend the notion of a root by defin- 
ing the nth root of a number, where n is any positive integer. 


12 ALGEBRAIC OPERATIONS [Cuap. I. 


DEFINITION. An th root of a given number is a number 
whose nth power is equal to the given number. 


For a given number z, the nth root is written ~/z. When 
n 1s 2 the root is called a square root. When vn is 3 the root 
is called a cube root. The number n is called the index of the 
root. 

From (38), we note that the nth root of a” is a”; for, a™ 


taken n times as a factor gives a". Thus, V28 = Va!-a! = at, 
Ayia = Wyia 4 = ya, 7/710 = x? W(x? — a?)t = x? — a? 
In general 
Van = an, 

This may be stated in words as follows: The root of a power is 
found by dividing the exponent of the power by the index of the 
root. 

16. Algebraic operations. An operation that involves only 
a definite number of the four fundamental operations and of 
the operations of finding powers and roots is often called an 
algebraic operation. 


EXERCISES 
Extract the roots indicated in the following : 

1. V/16a'. 5. ~/—8a'. 9. Vf ain”, 
2. \/40”, 6. V 16a. 10. x4. 

3. /4a™. 7. v/a. 11. V/a%?. 
4, ~/8a3. 8. v/a". 12. Yop. 


13. Expand (a — b)? and check the result when a = 2, b = 1. 
14. Expand (a — b) and check the result by making a = 8, 


MISCELLANEOUS EXERCISES 
Simplify the following by performing indicated operations : 
1. 1—[-1 + {25ers 6)} — 1]: 
2. 4a — [—-6y — (5 — 2y — 3x) + 4a] — [5 — (4y — 32)). 


. Arr. 16] MISCELLANEOUS EXERCISES 13 

av af a - Se 

e+ y* —- ry 

Va+vVa 1 

4. eee —: Check result for x = 9, a = 4. 
Vzi-Va Vrt+vVa 

5. (a + b)[a? — (ab — b?)]. 

6m* + m*> — 29m* + 27m —9 Check result by making 


3. Check result by making x = y = 1. 











os 3m + 5m? — 7m + 3 m= 1. 
7 T3x* — 2.92° + 0.92? + ae 
E 0.3% — 0.5 
6x22" Qar2m “= 1 a3 oa, y® 
< : ; : 10) 6 eee 
2 3 ah Aes FH xe+auyt+y? 


falta =o, y = 1,2 = 0, and a = 1, find the value of : 
fee ie — (y+ 2) } — [er —(y —2 — a)]. 

12. Ife =m+n-—2p,y =m —2n+p,andz =n+p — 2m, 
show thatx+y+z2=0. 

138. If a=2°4+ 427-1, b=1 —2 — 22”, and c = 273 + 27? 
+ 2-+1, find the value of — a — [b — (2a — c) +c] whenz = 2. 





14. Find the value of he ea when a = 4, b = 10, 
andc = 4. : 

15. Find the value of eee oe when a = 4, b = 10, 
and c = 4. 


16. Divide x° — y® by x’ — y’, and check the result by mak- 
ing « =2 and y=1. Why can not the check be made for 
ee ey eae Wy 

17. Divide 7 — y™ by x” + y™. 

18. Divide wR? — mr? by r(R — 7). 

19. Divide 47R? — 47r° by R — r. 

20. Divide §7D* + ¢7d3 by 7(D + d). 


- Historical note on the fundamental laws. When the attention of the 
student is first called to these laws, it is not unlikely that he will regard 


14 ALGEBRAIC OPERATIONS | [Cuap. I. 


them as self evident propositions. That they are not self evident should 
be clear from the fact that systems of numbers have been devised, that do 
not follow the same laws of combination as apply to the numbers we use 
in algebra. In one prominent system, we have the product ij = — ji. While 
the study of kinds.of numbers that do not obey the commutative, associa- 
tive and distributive laws, belongs to advanced mathematics, the fact of 
the existence of such numbers is mentioned here to emphasize the fact 
that the laws of algebra are assumptions, even if they have a wide applica- 
tion in experience. Strange as it may seem at first thought, we find, upon 
examining the history of algebra, that much was known of certain complex 
processes before it was understood that laws about order and grouping are 
at the basis of these operations. Sir William Rowan Hamilton (1805-1865) 
stands out prominently as a man who helped to make clear the nature of 
algebraic processes by devising numbers that do not obey all of the laws 
given in this book. 


CHAPTER II 


LINEAR EQUATIONS IN ONE UNKNOWN 


17. Equalities. A statement that one expression is equal 
to another expression is called an equality. The two expres- 
sions are called the members of the equality. 

There are two kinds of equalities, identical equalities or 
identities, and conditional equalities or equations. 

The two members of an identity are equal for all values of 
the symbols for which the expressions are defined.* Thus, 


xv? — a? = (4 — a)(x + a), 5a = 10a — 5a, 


are identities. 


But in the equality 
x—5 =4, 


the two expressions x — 5 and 4 are equal only whenz = 9. An 
equality of this kind, in which the members are equal only for 
particular values of the letters involved, that is, are not equal 
for all values, is sometimes called a conditional equality. In 
this book, we use the term equation to mean conditional 
equality. When it seems necessary to emphasize that an 
equality is an identity and not a conditional equality, we 
use the sign = instead of the sign = between the members. 
But the sign = will be used for both identities and equations 
when this usage can lead to no confusion. 


* This statement implies that we may not assign values to the lctter 
1 =e z 

1-2 1-2 

excluded when x = 1, since division by zero is excluded. (See Art. 14.) 


is 








which will make the members meaningless. Thus, 


16 LINEAR EQUATIONS IN ONE UNKNOWN  [Cuap. II. 


EXERCISES 
Which of the following equalities are identities? 
1. (x — a)? = x? — 2ax + a’. 





2. sine lth pt 

Tio 
8.27 — 32-4 2es' 0, 
4, 27°+6=0. 

2 pa ve 
5. Top teh ieee 
6. 472+ 474+ 1 =0. 


18. Equations and identities as sentences. An equation 
may be regarded as essentially an interrogative sentence, ask- 
ing for values of the letters that make the members equal. 


Thus, 
x-—-5=6 


asks for the value of x that makes z — 5 = 6. 

An identity is essentially a declarative sentence, stating the 
fact that two members are equal without regard to the values 
of the letters. Thus, 


x? + 2ax = 2x? — x? + 8ax — ax 
for any values that may be assigned to the letters. 


19. Solution or root of an equation. To solve an equa- 
tion in one unknown is to find values of the unknown that 
reduce the equation to an identity. Any such value is said 
to satisfy the equation and is called a solution or root of the 
equation. 


20. Equivalent equations. Two equations are said to be 
equivalent when they have the same roots; that is, when each 
equation is satisfied by the solutions of the other. Thus, the 


equations 
¢ — Dee and 37 —- 6 = 0 
are equivalent. 


Art. 21] OPERATIONS ON EQUATIONS 17 


21. Operations on equations. It is here taken for granted 
that in the first course in algebra the student has made con- 
siderable use of the following operations on members of an 
equation in the process of finding solutions : 

(1) Adding the same number.to both members 

(2) Subtracting the same number from both members. 

(3) Multiplying both members by the same number, other 
than zero. 

(4) Dividing both members by the same number, other than 
zero. 

These operations are permissible because they lead to equiva- 
lent equations. The operations (1) and (2) are often replaced 
by an equivalent operation called transposition. It consists 
in changing a term from one member of an equation to the 
other and changing the sign of the term. 

- The necessity for the restriction in (3) that the multiplier 
is not zero may be seen from the following examples : 


Example 1. Given z—1 =3. (1) 
Multiply by x — 2, x? — 3% +2 =3(a — 2). (2) 
From (2) x? -—-6x +8 =0. (3) 
Whence, (x — 2)(a@ — 4) = 0. (4) 


But, x = 2, and x = 4 satisfy (4) while (1) has the root 4 only. 


The non-equivalence of (1) and (4) is due to the fact that the multi- 
plier  — 2 is zero when x =2. Such values as x = 2 which satisfy a cer- 
tain derived equation, but do not satisfy the original equation, are 
sometimes called extraneous roots. 


Example 2. Given 2? — 3x +2 = 3x —- 6. (1) 
Divide by x — 2, z—1 =3, . (2) 
and x =4. (3) 


But (1) has solutions x = 2 and x = 4, while (2) has only the solution x = 4. 


The non-equivalence of (1) and (2) is due to the fact that we have made 
the error of dividing formally by x —2, which is zero when x = 2. But 
division by zero is excluded from algebraic operations. (See Art. 14.) 


18 LINEAR EQUATIONS IN ONE UNKNOWN  [Cuap. II. 


22. Linear equations in one unknown. An equation of 


the form 7 
ax+b=0,a40 (1) 


is called a simple or linear equation in the unknown xz. We 
have solved many such equations in the first course. 
The general solution of (1) is 


as may be verified by substitution, but it is in general better 
to treat each case separately than to remember the formula. 


2%. Verification by substitution. The operations of Art. 
21 are useful in finding solutions, but the solution is incom- 
plete until the values of the unknown are substituted in the - 
equation to be solved, and are shown to satisfy it. 


EXERCISES AND PROBLEMS 


Solve the following equations for x: 
1. 32 + 6 = 524 4+ 10. 


SOLUTION : 3a +6 = 54 + 10. (1) 
Transpose and collect, —2x¢ = 4. (2) 
Divide by —2, z= 2. (3) 


CuEck. Substitute in (1), and 
3(-2) + 6 = 5(-2) +10. 


0 =0. 
jae Seer 
ieee eed 
oy or re ; 
SOLUTION : Oy he 7 ema (1) 
Multiply by 12, 6z +42 = 15 — 122. (2) 
Transpose and collect, 22x = 15, (3) 
15 
© = 55 (4) 


Art. 23] EXERCISES 19 


CHECK. From (1) and (4) 
15 5 Woy ko 
44° 99 4 99? 
25 25 
44 44 
3. (x2 +9 = 2x — 6. 
4. 54+7 = 2x7 + 10. 
§. 54+ 7 = 27 + 9. 
Gor 27+ 1) = 6r?.—-9. 
ri 
8 
2 


or 


~ d(@ +1) + 2? = x? + 12. 
moe 2) = 3(c¢ + 1) — 13. 
me + 1)(z +3) =a(e — 2). 
10. ax +a = 6a. 
iv 3(a — xz) = 9a. 
12. 2(a? — 27 + 2) — (a +1) = 22? + 10. 
13. ‘(@ + 2)(z@ — 5) = («1 + 4)(2 — 1). 
ia. 2a + a(x — 1) + 1] = Qe —1)(x + 2). 
15. (m+ n)x + (m — n)x = mn. 
$4 aa, 
x+b 
b 
Tooeeod + o.00 = 2.52 -+- 17.5. 
19. 0.1(52 + 20) — 0.4(a — 5) = 25. 
5 —8 2(10 — 2) 
ASA Teak aera ge 





pe ol. is 
a 


22. p(p —x) = yy +2). 
23. mx = mW + MWe — nx.* 
24. y—5 = m(x — 10). 


* The subscript notation is found very convenient both in pure and in 
applied mathematics. The symbol m is read ‘“‘m sub 1.” 


20 LINEAR EQUATIONS IN ONE UNKNOWN  [Cuaap. II. 


The following exercises, 25-32, give relations in the nota- 
tion of physics. 

25. Given s = vt, solve for v. 

26. Given s = vt + so, solve for v. 

27. Given s = $gt?, solve for g. 

28. Given s = 3gt? + so, solve for g. 

29. Given EH = 4Mv’, solve for M. 


I 





30. Given D = Ww » solve for W. 
UW — WW 


81. Given F = 2c + 32, solve for C. 


82. Given PV = po( 1 + =a) express ¢t in terms of P, V, p, v. 

33. What extraneous root is introduced into an equation 
x = 5 if both members are multiplied by 2? by a2 — 2? by 
x+3? 

34. What extraneous root is introduced into the equation 
ax +b =0 if both members are multiplied (1) by x? (2) by 
x—2? (8) byx-—a? 

35. What root is lost if the members of x? — 4 = x — 2 are 
divided by x — 2? 

36. If each member of an equation is multiplied or divided — 
by the same expression that does not contain the unknown, 
are the roots changed ? 

37. Five times a certain number when diminished by 23, is 
equal to 5 more than the number. What is the number? 

38. A father is four times as old as his son and the sum of 
their ages is 65. What is the age of each? 

39. Find three consecutive integers whose sum is 78. 

40. What number must be added to each of the numbers 
18, 108, and 48 in order that the product of the first two sums 
shall be equal to the square of the last sum? 

41. What number must be added to each of the numbers 
a, b, and c, in order that the product of the first two sums 
shall be equal to the square of the last ? 


Art. 23] PROBLEMS 21 


42. How soon after noon will the hands of a clock be to- 
gether again? 
SuaceEsrion: Let x = number of minute spaces which the minute hand 


has traveled from noon until it overtakes the hour hand. Then 7 will be 


the number of spaces the hour hand covers meanwhile. The difference, 


xL— 5 is 60 spaces. Why? 


43. Denochares has lived a fourth of his life as a boy; a 
fifth as a youth ; a third as a man; and has spent 13 years in © 
his dotage. How oldis he? (From the Collection of Problems 
by Methodorus, 310 4. D.) 

44, A man can do a piece of work in 4 days, another in 6 days - 
and a third in 12 days. How many days will it require all to 
do it when working together ? 

45. A room is 4 feet longer than it is wide, and if the length 
were decreased by 2 feet and its width increased by 4 feet, the 
area of the floor would be increased 40 square feet. Find the 
dimensions of the room. 


PROBLEMS PERTAINING TO MIXTURES 


46. A grocer mixes two grades of coffee which cost him 20 
cents and 35 cents per pound respectively. How much of 
each must he take to make a mixture of 60 pounds which he 
can sell at 30 cents per pound with a profit of 20%? 

47. A dealer has 2000 gallons of alcohol 85% pure. He 
wishes to add water until it is 75% pure. How much water 
must he add? 

48. How much cream that contains 25% butter fat should 
be added to 1000 pounds of milk that contains 34 % butter fat 
to produce a standard milk with 4% of butter fat? | 
— 49. How many pounds each of water and milk with 4% of 
butter fat should be taken to give a mixture of 160 pounds 
with 31 % butter fat? 

50. If 24 pounds of iron weigh 22 pounds in water and 20 
pounds of lead weigh 19 pounds in water, find the amounts 


22 LINEAR EQUATIONS IN ONE UNKNOWN [Caap. II. 


(weights in air) of iron and lead in a mass which weighs 180 
pounds in air and 170 in water. 

51. A dealer mixes a pounds of tea worth x cents a pound 
with 6 pounds of tea worth y cents a pound and with c pounds 
of tea worth z cents a pound. Find the value, v, of the mix- 
ture in cents per pound. 


PROBLEMS PERTAINING TO BUSINESS 


52. The gross income of a certain man was $110 more in the 
second of two years than in the first, but in consequence of an 
income tax of 1% on the part of the income above $3000, the 
net income (after paying tax) was the same in the two years. 
Find his income in each year. 

53. A married man has a $4000 exemption from income tax, 
but pays at the rate of 1% on the rest of his income. He © 
finds that after deducting income tax, his actual income is 
$5980. On what amount does he pay the 1% tax? 

54. A man made two investments amounting together to 
$6000. On the first he gained 10%, and on the second he 
lost 5%. His net gain on the two was $60. What was the 
amount of each investment ? 

55. At a ball game the charge was 75 cents for each re- 
served seat and 50 cents for each general admission ticket. 
The ticket seller found he had sold 4320 tickets for $2785. 
How many people bought reserved seat tickets ? 


PROBLEMS INVOLVING THE LEVER 


Two boys, A and B, are balanced on a teeter board as shown in Fig. 2. 
The board is twelve feet long 
with the point of support at the 
middle. The boys find that they 
must sit at distances from the 
Fie. 2 support such that the products ob- 
; tained by multiplying the weight 
of each by his distance from the fulcrum or point of support are equal. : 


Thus, (78) (5) = (65) (6). 





6-ft: 








ArT. 23] PROBLEMS 23 


In the general problem of the lever the force F required to lift the weight 
W depends upon the position of the point of support (fulcrum) and the 
lengths of the arms of the lever. If f and w represent the distances of the 
force and weight respectively from the fulcrum, then the law of the lever 
is expressed by, 








Fig. 3 


56. A and B weigh 70 and 120 pounds respectively. On 
a teeter board, A is to sit 6 feet from the fulerum and B 5 feet. 
A stone is placed on A’s side 5 feet from the fulerum so that 
the sides will balance. How heavy is the stone? © 

57. A and B together weigh 320 pounds. They balance when 
A is 8 feet from the fulcrum and B is 6 feet. Find the weight 
of each. 

58. How heavy a stone can a man, by exerting a force of 
200 pounds, lift with a crowbar-6 feet in length if the fulcrum 
be six inches from the stone (neglect weight of crowbar) ? 


PROBLEMS INVOLVING UNIFORM MOTION 


The rate of uniform motion is the distance traversed in each unit of time. 
The rate of motion is often called the speed or velocity. .It is customary 
to use s for distance or space traversed, v for velocity, and ¢ for the time. 


59. A Zeppelin flies 55 miles per hour and an aeroplane 70 
miles per hour. If the aeroplane is 7.5 miles behind the 
Zeppelin, how long will it require to overtake the Zeppelin? 

60. A freight train leaves New York for Chicago making a 
speed of 30 miles per hour. Five hours later a limited express 
leaves New York making 54 miles per hour. In what time will 
the express overtake the freight? : 


24 LINEAR EQUATIONS IN ONE UNKNOWN  [Cuap. II. 


61. In how much time does the minute hand of a clock gain 
a complete revolution on the hour hand ? 
62. If the speed at which sound travels in air is given by 


vy = 1090 + 1.14(¢ — 32), 


where v is the speed in feet per second, and ¢ the temperature 
of the air expressed in Fahrenheit degrees, find the tempera- 
ture of the air when sound travels 1120 feet per second. 


CHAPTER III 
FACTORING 
24. Rational integral expressions. By a rational integral 


expression in certain letters, say in x, y, 2, we mean the sum 


of terms of the type axmyn2?, 


where the exponents m, n, and p, are any positive integers. 


Thus, 523 + 10a? + 42 — 6, 
Bry +y?V5 — a0? — «v7, 
and ax’y + 10xy? — cxyz 
are all rational integral expressions in all the letters involved, while 


3x , . : : ‘ 
5, are rational integral expressions in x but not in y. 
Y 


323 + avy, and x? + 
It should be noted that a rational integral expression with 
respect to a letter x contains no indicated root of that letter, 
as Va, and does not have that letter in the denominator of a 
fraction. 

The degree of a term of a rational integral expression is the 
sum of the exponents of the letters in the term. The degree 
of a rational integral expression is defined as that of a term 
whose degree is equal to or greater than that of any other term 


in the expression. 






Thus, the degree of 3x?y + 


We are especially concern hapter with the factors 


of rational integral expressions. 

25. Definition of factoring. We shall use the term factor 
to mean a rational integral factor. 

A rational integral expression is prime when it has no factor 
besides itself and unity. : 


26 FACTORING [Cuap. III. 


Finding the prime factors of a rational integral expression 
is called factoring the expression. 


26. Important special products. Factoring in elementary 
algebra depends mainly upon the recognition of certain type 
products. ‘The following have already been treated in the First 
Course. They should be reviewed and memorized. 


a. Common monomial factor. ax +ay =a(x+y). 
Example. 2am — 4a? = 2a(m — 2a). 
b. Difference of two squares. a? — b? = (a+ b)(a — b). 
Example. 812? —1 = (9x + 1)(9a — 1). 
c. Trinomial square. a? + 2ab + b? = (a+ b)*. 
Example. 42? — 12x%y + Oy? = (2x — 3y)?. ‘- 
d. Trinomial of the form 2? + (a + b)a + ab = (a +a)(x+5). 
Example. 2? + 7x +10 = (4 + 5)(a@ + 2). 
e. Sum of two cubes. a’? + 6? = (a + b)(a? = ab 4+ B?). 
Example. 1252° + 8y!5 = (5a? + 2y5) (254 — 10x?y*> + 4y). 
f. Difference of two cubes. a? — & = (a — b)(a? + ab + B?). 
Example. my? — 27n® = (my — 3n') (my? + 38mn3y + On‘). 
g. Factors found by grouping. 

ax + ay + ba + by = (a TF b)(a + y). 


Example. 
2ax + 4bx — 8ay — 6by = 2x(a + 2b) —3y(a + 2b) = (Qe — 3y) (a + 2b). 







h. Trinomial of form +c. Certain expressions of 
this form can be factored ection. 


Example. Factor 6x? + x — 10. 


SotuTion. The factors are two binomials whose first terms are 3a and 
2x or 6x and x, and whose last two terms are +3 and +5, or +1 and #15, 
if the coefficients are integers. We must now choose the terms of the bi- 
nomials so that the algebraic sum of the cross products is +x. By trial _ 
we find that the factors are 2x — 3 and 3x + 5. 


_ ART. 26] EXERCISES IN FACTORING 


EXERCISES 
Factor the following : 


a+ ab. 

3+ 3x. 

m* — A, 

a? + 10a + 25. 

m* — 9m + 14. 

a’bt — x4. 

mp +mq+np + nq. 
10am + 8aq + lima + 12qz. 
5r3 + 5. 

k? + 6k — 55. 

. ct — 12c? + 36. 

. op? — 9p — 18. 

. Gar? + 24ax? + 24az. 
- Zax + 6a? — 4ay. 
moe l. 

. re 4+ 22r + 117. 

i 


: pepe S. 
er 


ain a ere glad) te 


a ee ee ee ee 
oOnPr wo DH OS 


= 
~j 


= 
ioe) 


1 
Ve Sale 
serie te: 4 


. 3l? + 161 — 35. 

. am+bm+a+ob. 

. a + 20a" +100. Find two factors only. 
eee? — ye tl He. 
. 2 — 52" + 42". Fmd three factors only. 
. a —1. Find two factors only. 

ae 3. 

. abt+at+bd+1. 

. dax — Sby — day + 36a. 

Aap rd Oe : 


meowene Fe 
woe Oo © 





NNN NY NY WD 
on on -» Ww 


28 


29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
46. 
47. 
48. 
49. 
50. 
51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 
59. 


FACTORING [Cuap. III. 


mt — ni. 

(38a — y)(2a + 4) + 5(8a — y). | 
(2a — 3b)? — 4b?. 

a” — b? + 2be — c?. 

x? + 4x? + 42. 

a? — Tab + 12b?. 

8x3y® — 125m*. Find two factors only. 
3bmx + 2bm — 3pqu — 2pq. 

1 — x. 

60a? + 8ax — 32”. 

a‘b + 5a%b? + 40763. 

24miny — 81lmy*. 

(m + 2)? — 5(m + 2) — 176. 


5a? — 8ab — 4b?. 
1z° — 12° — 842. 
x? — 4ax — 4b? + 8ab. 


(a + b)(c? — d?) — (a? — b?)(c — d). 

2a? — 5x — 38. 

2ilx? — 232 + 6. 

(Qx +a —b)?— (x —a+b)?. 

x — 2° — 647? + 64. Find six factors. 
x® + 1623 + 64. 

12x? + 25a — 7. 

152* — 2° — 62. 

9a2x* — 30a7x? + 25a72?. 

(x — 1)? — 9(@ — 1)? + 8(@ — 1). 

1 — a — 2ry — y?. 

a*c? + acd + abe + bd. 

2? — 64" +8. Find two factors only. 
a‘? — 5a?Pht — 146%. Find two factors only. 
a"b2n+2 — 16a°"b?c%, Find four factors. 


Arr. 27] TYPE PRODUCTS 29 


_ 2%. Further type products. The following type products 
are also useful and should be mastered. 
i. Square of a polynomial. The square of a polynomial 
equals the sum of the squares of the terms plus twice the product 
of each term by every term that follows. Thus, 


(a+b+c+d)?= 
Beeb ce + d* + 2ab + Zac + 2ad + 2be +- 2bd + 2cd. 


If a polynomial can be put into the form of this product it can 
be factored. 
Example. 
x? + 4y? + 922 + 4ry + 6rz + 12yz 
= (x)? + (2y)? + (82)? +2-2-2y4+2-4-38242- 2y: 3z 
= (x + 2y + 32). 
j. Cube of a binomial. By performing the multiplication, 
we find that 
(a + b)® = a? + 3a7b + 8ab? + 0b. 

If a polynomial can be put into this form it can be factored. 


Example. 82° + 362*y + 54ry? + 27y' 
= (2x)' + 3- (2x)? (8y) + 3- (2x)- By)? + (By)? 
= (22 + 3y)’. 
The student should find the cube of a — 6. 
k. Expressions that can be reduced to the form of the 
difference of two squares. 


Example 1. Factor x* + xy? + y'. 
By the addition and subtraction of x?y? we have, 
= (x2 + y?)? — 2%? 
= (2? + y? + cy) (2? + y? — ay). 
Example 2. Factor 36a‘ + 44ab? + 25b*. Find two factors only. 
If 16a2b? be added to 36a! + 44a2b? + 25b4, the resulting expression is 
a perfect square. Adding and subtracting 16a7b?, we have 
36a + 44a%b? + 25b! = 36a! + 60a2b? + 25b4 — 160%? 
= (6a? + 5b?)2 — (4ab)? 
= (6a? + 5b? — 4ab) (6a? + 5b? + 4ab). 


30 FACTORING [Cuap. III. 


EXERCISES 
Factor the following : 
x? + y? + 1627 + 2xy + 8x2 + Byz. 
4a? + 9b? + c? — 12ab + 4ac — 6be. 
x? + 6x7y + 12xy? + 8y’. 
8m? — 36m2n + 54mn? — 2773. 
ed ral re ie rhe a 
zt — 3274+ 1. 
a* + 4a? + 16. 
a® — 6a’b + 12ab? — 8b’. 
x y%z3 + 6aroytz? + 1273y2z + 8. 
x* — 11lxz? +1. 
x? + Dy? — bry + By — 2a 4+ 1. 
~ d+ 32? + 32%. 
x§ — 321 + 32? — 1. 
~ cP + yt + Qr3y? + Qy2z + 2? + Qrz. 
iets: fe ee 
. 2+ 2e3 4+ 32? + 274 4+1. 
a’z® — 3a*x* + 3a*x? — a’. 
4a‘ + 12a°b + 12a7b? + 4ab?. 
8as™ — 12a?2"5" + 6a™b2" — 5%. 
aim +o?" + 1. Find two factors only. 


— 
Fe Se ee ls eee bo ee 


XE ll eal ool oe = 8 
SOMABRAR WD 


. Factor theorem. Divide 3x7 — 10% +7 by 2 - a. 


cS) 
@ 


32? — 10% +7 |. Dee 
ox? — 3ax ox + 3a — 10 
(3a — 10)a +7 
(8a — 10)x — 3a? + 10a 
3a? — 10a +7 





The remainder is seen to be the same as the dividend when 
a is substituted for x. This exercise illustrates the principle 
that. the remainder resulting from dividing a rational integral 
expression in «x by « — a may be obtained by substituting a for x 
un the given expression. 


ArT. 28 ] FACTOR THEOREM dl 


To find the remainder when the above expression is divided 
by x — 1, we substitute 1 for vz. This gives 


o:1l?—10-1+7=0. 


Since the remainder is zero, the expression is divisible by x — 1; 
that is, c — 1 is a factor of 3x? — 10a + 7. 

This exercise illustrates the important principle known as 
the 

Factor THEOREM.* If a rational integral expression in x 
becomes zero when a is substituted for x, then 2 — a isa factor of 
the expression. 

The factor theorem gives a simple means of factoring cer- 
tain expressions. 


Example 1. Factor 5x? — 13x + 6. 


We wish to find a factor of the form x —a. In order that the given 
expression be exactly divisible by x —a, where a is an integer, a must be 
a factor of 6, that is, a must be +1, +2, +3,or +6. If we substitute 1 for 


x we have, 
5-12-18-146 = -2. 


If we substitute —1 for x we have, 
5 - (-1)? —13- (-1) + 6 = 24. 
If we substitute 2 for x we have, 
5-2? -—-13-2+46 =0. 


Hence x — 2 is a factor of the given expression. The other factor is found 
by division to be 5x — 3, and we have 


5a? — 134 +6 = (x — 2) (5a — 38). 
Example 2. Factor 3x’ + 4¢ ~15. 
By substituting —3 for x we have, 
| ees ep 4. (3) — 15 - 27 —12 = 15 = 0. 
Hence x — (—3) =x +3 is one factor. The other factor is found by di- 
vision to be 3a — 5, and we have 
3a? + 44 —15 = (x + 3) (82 — 5). 


* For the expression az? + bx + c, a proof is given on p. 130. For proof 
of the general theorem see Rietz and Crathorne, College Algebra, p. 121. 


32 FACTORING [Cuap. III. 


Example 3. Factor x* — 2x7? — 52 + 6. 


By the use of the factor theorem one factor of the given expression is 
found to be x —1. The quotient found by dividing 2? — 2a? -— 544+ 6 
by «x —1lis x? —x-—6. The factors of this quotient are x + 2 and z— 38. 


Hence 
ze — 22? ~ 54 +6 = (& —1) (44+ 2)@—s) 


EXERCISES 
Factor the following expressions : 
1. x? — 34 4 2. 5. 2° + 3a? — 5x — 15. 
2. 377 — 9x + 6. 6. 2? — 12%7 + 27x + 40. 
3. 2x3 + 347+ a. 7. 2 + 8a? + 17x + 10. 
4. 2° + 627? + 32 — 10. 8.004 = 27a 
9. e*+2°+4+ 8 +8. 


Show without actual division 

10. That a‘ — 6! is divisible by a—b and by a+b. 

11. That «2° — 5xty + lla*y? — 147?y? + 9ry* — 2y® is divisi- 
ble by x — y. 

12. That a> — 6° is divisible by a — b, but not by a + b. 

13. That a® + 6° is not divisible by a + b ora — 6. 


29. The sum or difference of the same two powers of 
two numbers. 

The type form is 

a” = Db”. 

By applying the factor theorem, we can determine whether 
a+bora-— bis a factor of a" + 6". 

1. Divisibility by a — b. 

Let n be either an even or an odd integer. If we substitute 
a = b in a” — b", we have 6” — b” = 0;if we substitute a = b 
in a” + 6" we have b” + b” = 2b”. 

Hence a — b is a factor of a” — b", but not of a” + b”. 


Arr. 29] FACTORING OF a" 33 


2. Divisibility by a + 6. 

Let n be odd. Substituting a = —b in a” — 6" we have 
(—b)" — b” = —2b"; substituting in a* +6" we have 

(—b)" + b” = -b" + 6" = 0. 
Therefore if n is odd,a + isa factor of a” + b", but not of 
a” — 6". 

Let n be even. Substituting a = —b in a” — b” we have 
(—b)" — b” = 6” — b" = 0; substituting in a” +6", we have 
(—b)" + b” = 6" + b* = 26". 

Therefore of n 1s even, a + 6 is a factor of a” — b”, but not of 
a” + b*. 
| Example 1. Factor a’ — b®. Find two factors only. 

SoLuTion : a> — b5 = (a — b)(at + ad + ab? + ab? + B4). 

The second factor is found by division. 

Example 2. Factor a’ +b’. Find two factors only. 

SoLuTIon: a? + b7 = (a +b) (aS — ad + ath? — a3 + abt — ab + bS), 
: Example 3. Factor a! — b!. Find four factors. 

SouuTion : a!° — b! = (a> — b)(a5 + 55). Finish the solution. 
Example 4. Factor x!° + 243y5. Find two factors only. 
SOLUTION : 
gid | 243y° = (x?) 5 4. (3y)> 
= (x? + 3y)[(@)* — (w)83y + (x?)? (By)? — 27(3y)* + (By)4] 
= (22 + 3y) (28 — 3a°y + Oaty? — 272?y? + 81y*). 


The above examples illustrate the following rules which are 
useful when we wish to find the quotient of a” + b” divided by 
a+borbya-6. 

When the divisor is a — 6 all the terms of the ‘quotient are 
positive. ! 

- When the divisor is a + b the terms of the quotient are alter- 
nately positive and negative. 

The number of terms in the quotient is n, if the divisor is a 

factor of the dividend. 


34 FACTORING [Cuap. III. 


The exponent of a in the first term is n — 1 and decreases by 
1 in each succeeding term. 

The exponent of b is 1 in the second term and increases by 
1 in each succeeding term. 


EXERCISES 

' Factor: 
1. “2° + 7°. 
Zaye Find two factors only. 
3. ty", Find. three factors only. 
4, x6 + y?, Find two factors only. 
5. 32a° — 1. Find two factors only. 
6. v4 + y’. Find two factors only 
Ton Ce Oe Find two factors only. 
ioerate eg pe Find six factors. 
9. 12827 — y#. Find two factors only. 


a 
ee 


243a" + m>. Find two factors only. 


. Find the first four terms in the quotient = S ep 


— 
= 





12. Find the first four terms in the quotient e ges 





13. Find the remainder when x* — 7x* + 2° — 547 —247 1s 
divided by x —1; by x +1. 

14. Find the remainder. when 42° — 224 — 773 +249 is 
divided by « — 2; by x + 38. 


30. Summary of factoring. The following suggestions will 
be found helpful in factoring. 

I. Take out monomial factors, keeping in mind numerical 
factors. 

II. After the monomial factors have been removed, if such 
factors occur, see if the expression to be factored can be classed 
under any of the following type forms of this chapter: 


‘Arr. 30] SUMMARY OF FACTORING 35 


1. Binomials. 
(a) The difference of two squares, as a? — b?. 
(b) The sum or difference of the same two powers, as 
Ge b. 
2. Trinomials. 
(a) Trinomial squares, as a? + 2ab + b?. 
(b) Trinomials of the form x? + (a + b)a + ab. 
(c) The general quadratic trinomial, ax? + bx + ¢. 
3. Polynomials of four or more terms. 


(a) Expressions to be factored by grouping terms. 
(b) Difference of two squares or expressions that can 
be reduced to that form. 
(c) Square of a polynomial. 
_(d) Cube of a binomial. | 
_ III. See if a rearrangement of terms will bring the expres- 
sion under any of the above forms. 
IV. If the above methods fail, try the factor theorem. 
V. Test each factor to see if it can be further factored. 
VI. It is convenient to remember that x? + y?, v7 4+ xy + y’, 
and x? + zy — y? are prime. 


MISCELLANEOUS EXERCISES 


Factor the following expressions : 


1. x3 + 4a? + 42. 3. 162° + 250y°. 
2. x2 +-(a — b)x — ab. 4. mx — nx + pr. 
5. 9(2a — d)? — 4(3a — x)? 

6. (4a — 3b)(7m — 2p) + (a + 4b) (7m — 2p). 

7. 8a? — 2lab — 9b’. 

8. ax — be + cx + ay — by + cy. 

9. 8m — 36m?q — 27q° + 54mq’. 


10. 16a*x* — 8a7z? + 1. 


36 


uli Be 
12. 
13. 
14. 
15. 
16. 
Uy 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 


FACTORING (Cuap. III. 


m? + 4n? + 1+ 4mn — 2m — 4n. 
a® + 68, 

m> — b>. Find two factors only. 
at + +. 

Gt 7 | 

8la? — 16(2a — 8z)?. 

x — y*, Find two factors only. 
gi — yy, Find three factors: 
x? — (c+ 5)x + Se. 
n(e-y)—-x+y. 

aba’ +4- 8abx? — abx — 8ab. 

30x? + 13x” — 77. 

a’ + 27a’. 

l2az%y! — 36a*bx?y? + 27a°b?x. 

x? + 5a? — 29x — 105. 
er>+y?+4+ 2ry — 4x — 4y. 

a‘ + 4. 


Hint: Add and subtract 4a?. 


28. 
29. 
30. 
31. 


32. 


33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 


x* — 10x? + 9. 

a” + lla” +18. Find two factors only. 
mtni+m+n. 

k10 — k. Find four factors. 


eetat+ i. 

p2nt4 = r2, 

8m'n  — 36m?2n2x + 54mnx? — 2723, 
vit 8a? +a —- 42. 

1224 + 962. 

62? + a — 2. 

5a? + 2a — 3. 

a? — 4ac + 4c? — 2. 


- Art. 30] 


41. 
42. 
43. 
44, 
45. 
46. 
47. 
48. 
49. 
50. 
51. 
52. 
53. 


54. 


55. 
56. 
57. 
58. 
59. 
60. 


61. 


62. 
63. 
64. 
65. 
66. 
67. 
68. 
69. 
70. 


EXERCISES IN FACTORING 


a’” — 3a" + 3a" — 1. 

m> +1. Find two factors only. 

582 — 1]. Find two factors only. 

2% — 7, Find two factors only. 

xe — 3x4 + 2. 

10° + 1000. Find two factors only. 
at + 464. 

3bmx + 2bm — 8anx — 2an. 

b? — 4b? + 8. 

a? — b? — 8ab(a + BD). 

(m +n)? — 3(m + n)*p + 38(m + n)p? — p’. 
m+7+(m+7)?. 

a’ + 19a? — 216. 


30 30° 
Pee x? +”. 


32° — 1502° + 1472. 

vt + 203 + 3x7 + 2a 4+ 1. 

xg? — 54x —171. 

2a* + 102? — 25227. 

or +7 +9"-+1. Find two factors only. 
x -_ = + 4. 

px” — 2pgrry” + gy". Find two factors only. 
23 — 3%. Find two factors only. 

125¢!* + 752 + 1525 + 1. 

56x? — 38x + 6. 

Ibe — 1)? — 2(k — 1) — 8. 

(7 — 2)§ + 2(r — 2)?. 


a + yy’. 
r? — 5°, 
1 — (4.5). Find two factors only. 


37 


38 FACTORING [Cuap. III. 


31. Solution of equations by factoring. ‘The method of 
solving quadratic equations by factoring (Art. 84, First 
Course) can be extended to equations of higher degree than 
the second. 


Example. Solve the equation 32° + 16x? + x — 20 = 0. 
Factoring the left member, we have 
(x — 1)(@ + 5) (3% + 4) =0. 
We next set each of these factors equal to zero and solve for x and obtain 
x=1,2 = —5, andz=-— = These values check when substituted for x in 


the original equation. 


EXERCISES 

Solve the following equations and check : 

1. x? + 8x — 209 = 0. 5. v2 — da79-4+5=0. 
2. 36a" + 17 — > = 0; 6. 2° — 77 +6 = 0. 

3. 2° +422 +2-6=0. 7. 2° — 8 = 0: 

4. 102° — 112? — 62 = 0. 8. 22 +2? = 0. 

9. x* — 9277 + 8 = 0. 

10. x* — 102? + 35x? — 50x 4+ 24 = 0. 


11. 2? — 5ax — 24a? = 0. 14. 22° :— (a? 4 Jee 
12. x? = 16. 15. x* — 2a*z7? + a* = 0. 
13, 2 — 2? - 2420; 16. 2° —- 422527 


17, 32°.— 7° — 32-2 = 0. 
18. 2 — 674 lla —6=0. 
19. 2? + 5x? — Qe = 10. 20. 3021 + 1323 = 7727. 


32. Highest common factor. A number or expression 
which is a factor of two or more expressions, is called a 
common factor of those expressions. 

The product of all the common prime factors of two or more 
expressions is called their highest common factor (H.C.F.). 


ArT. 32] HIGHEST COMMON FACTOR 39 


Two expressions which have no common factor are said to 
be prime to each other. 

The H.C.F. of two or more expressions can be found by 
resolving eech of the expressions into its prime factors, and 
then finding the product of the common prime factors. 


EXERCISES 


Find the H.C.F. of each of the following sets of expressions : 


ve 


eS eh ee 
Pwd HE oO 


nth baal Sie tonsa ai dhe” had 


ax”, 2abx, and 3a*b?. 

12abc, 28a7bc?x, and 4a‘b*cy. 

60, 144, 84, and 156. 

52, 117, and 78. 

a? — b? and a? — 5ab + 407. 

x? + Qry + y?, 2? + xy, and x? — Txy — 8y’. 

240, 1260, and 3300. 

900, 1650, 3150, and 4850. 

12a? — 24ab, a? — 3b + 2b?, and 2a? — 3ab — 2b?. 


m2? — 3m — 4, 2m? — 7m? — 5m + 4, and m3 — 4m? — m +4 4. 


. 2 — y8, 2? + Qry + y*, and 24 — yt. 
.r?—6r+9,r? + Sr — 24, and r? — Or + 18. 

. a 4+ 8a2y + day? + y®, v3 + Qa?y + ry’, and xy + 2ry? + y?. 
. 4 —2?, and x? — 47 + 4. 


15. 1 — 2x, x? + 182 — 14, and 1 — 2’. 

16. 22 — (y+ 2z)?, (y — x)” — 2, and y? — (x — 2)?. 

17. ma+mb—mc, a?+b? +c? + 2ab — 2ac — 2bc, and 
(a + 6)? — c?. 

18. 23 — y3, x — y®, and 2° — y’. 


19. x22" — ryt, gen Qery" ag an and x22" — 1p 
20, —1)7, a — 1, anda? +2*-—<2z- Il. 


AO FACTORING [Cuap. IIT. 


33. Lowest common multiple. The lowest common multiple 
(L.C.M.) of two or more expressions has been defined to 
be the product of all of their different prime factors, each taken 
the greatest number of times that it occurs in any of the expres- 
sions. This definition gives the method of finding the L.C.M. 

It may be noted that the L.C.M. of two or more expressions 
is the expression of lowest degree which contains each of the 
given expressions as a factor. 


EXERCISES 


Find the L.C.M. of each of the following sets of expressions : 


8 


3ax?, a’bx, and 2ab2z. 

. 6m2n, —4mnz3, and 12mn?z. 

240, 225, and 500. 

160, 450, 600, and 840. 

z?24+ ary, 22+ y’, and x27 — 3xry — 4y?. 

a? — 14a — 32, a® + 2a, and 3a‘b. 
x+1,x2+2, and zx? + 32 4+ 2. 

a? + 7a + 10, a? + 6a? + 3a — 10, and a? + 4a — 5. 
m — 8m? + 16, 5m? + 10m, and 3m. 

ax — ay + bx — by, a? + 2ab + b?, and x? — zy. 

. a?,a — 1, (a — 1)?, and (a? — 1)?. 

. 2m? + 2m — 1, 8m3 — 4m + 1, and 2m — 3m + 1. 

. a(m — n)(m — p), b(p — m)(m — q), and c(q — m)(n — m). 
(2 ~ 3)(a — 7), (8 — x)(@ — 2), and (7 —HG =a 
. a+ 8, 8 — a’, and a? — 4. 


rad Ng? ire SG AP SE pel led 


a eS SS 


CHAPTER IV- 
FRACTIONS 


34. Algebraic fraction. An algebraic fraction is the indi- 
cated quotient of two expressions. Thus, 


a 
b 
means a divided by 6. 

35. Reduction of fractions. The form of a fraction may 
be modified in various ways without changing the value of the 
fraction. For different purposes in the treatment of fractions, 
we make use of the following principles which have already 
been studied in the First Course : 

I. The value of a fraction 1s not changed by multiplying or 
dividing both the numerator and denominator by the same number. 


That is, 


II. Changing the sign of either the numerator or the denomina- 
tor of a fraction is equivalent to changing the sign of the fraction. 


That is, —— = — 


III. Adding two fractions having a common denominator gives 
a fraction whose numerator is the sum of the numerators and 
whose denominator is the common denominator. 








That is, ) ee oY 
cee. C C 
Likewise, gee ae 2 


42 FRACTIONS [Cuap. IV. 


IV. The sum and the difference of two fractions are expressed 


b 
» respectively. 








a C 
We can reduce b and F; to a common denominator, since by I, 


a _ad ge = 2. 
b> bd ee Wood 
By III, we can complete the process. 


V. The product of two fractions is a fraction whose numerator 
is the product of the numerators and whose denominator is the 
product of the denominators. 


That is, Pee mers 


VI. To divide one fraction by another, invert the divisor and then 
multiply. 


That is, 


The reciprocal of a number is 1 divided by the number. Thus, 
the reciprocal of a is = aaa i is” “Of ous - Hence, to divide. 
by a fraction, multiply by its reciprocal. 

Propositions III and V, stated for two numbers in each ease, 
are readily extended to three or more numbers. 


36. Reduction to lowest terms. To reduce a fraction to 
its lowest terms separate the numerator and denominator into 
their prime factors and then cancel common factors by division. 
We know by Principle I (Art. 35) that this cancellation may be 
performed. 


Art. 36] REDUCTION TO LOWEST TERMS 

















Reduce oo to its lowest terms. 
7 x -8 (a — 2) (a? + 27 + 2) 
SOLUTION : (¢ 2) = @ = 2) 
4 zc? +27 + ve 
meget 2)? 
EXERCISES 
Reduce to lowest terms: 
48 ap 
1 aa Tre 
Ga Bx a® — 6? 
2. Seapear a 
a2 a® + b 
“ a? + ab 2 a? +B 
4 aa 
4. ee? 19, Mt = @. 
a* — ab ny + my — qy 
a—1 (m + n)3 
oe Se ses 
5a2b3c gen ani Te 
acne 2 ee ae 
! 25a°b2c : (a + y")? 
13 2a7x? — a®x — bat 
' 3 — 3az? + 2a%x 
14 v + y? +2? + 2ry — 2x2 — 2yz 
é x? + Qry + y? — 2? 
15 ae + 62? -+ lle + 6 
; eT — 6 
a+ 
28. @2@+i+(a¢e+1) 
17 (a — b)(c — d)(b — c) 





(a — c)(c — b)(a — b) 
1g, HWS - yy") 
"(x3 + 3) (x? + y?) 


43 


44 | FRACTIONS 
x — Ses 597 — 13° 








tu: eee eee 
gen + 3 a co 
20. gin +l _ pany , 
°1 20a?m2 — 5a’m?n4 
* 20a2m — 20a?mn? + 5atmn4 
(a? Eb O72 
a ae 
2nhn n 2n 2npn 
23. a2"b"c + abc?" + ab"c 
abe 
oa (a +b)? + 4(a +b)? + (aie 
(a + 5)? 


[Cuap. IV. 


37. Addition and subtraction. The method of addition 
and subtraction of fractions is given in Principles III and IV 


(Art. 35). 


That is, we reduce the fractions to be added or subtracted 
to a common denominator, and then add or subtract the 


numerators. 
1 2 
Example. Add ma ee and ln See ae 
SOLUTION : 
1 y 
@ —2)@-3) ' @-3)@-4 
2-4 24 —4 
“@-2)@-3)@-4 "@-2@-3)@-4 
3x — 8 





~ (2 —2)(z — 3) (2 — 4) 





EXERCISES 
Perform the following additions and subtractions: 
Comet, “YY 
aie 3 + 3" i j ee 
1. v- “Tia 
ee die pe 
Z a u b 5 5 


ArT. 37] EXERCISES 45 


24. 


25. 











eee te D = =D Cc 
ape meg Me Sa 
1 2 dr z 
Aaa oe wees ate 
Peo, or — 3 12 ye a 9 a—b 

5 2 ie Tia eee: 

1 a 8 

a ee aree hy Me Bye ATS 

7 les) 


a 
mee 14,a+b+7—1. 


ba — 4. fo oe 17 
x—2 z* — 52+ 6 














2 a 
Sha 2£+b 2?+(a+b2+ab 
a b ab 
mee cha (acc) 
8 5 om — 4 


; eee SO Om? = m8 





Coie + 1- a+ il a? — 1 








Se aS 
°a—-b b-a ; 
__ es Re ee 
~@—yy—z). @-z)\e-y) (y-—2)@-2) 
eee ey 
ae? — Sry + Gy? sy? — 4zy + 2? 

x+a on x+6 1 xr+e 
x?—(b+c)x+be x?-—(a+c)x+ac 2*?—-(a+b)x+ab 
x+1 - 1 
ou Ss ee 


46 , FRACTIONS | [Cuap. IV. 
ant + b™ > a" — pb" 








26. ae, bn a qa” + bn 
27 3a a a—1 au 2 
"“@+ta—20 a*?—15a+44 °- a? — 6a —55 


u 


20.8 : : 











29. 1+-4+54+54+ 5+ 





all i iy: CR 
Sy ee 
31. l-y-y Peay 


32. By how much is the value of the fraction = changed by 


3 
adding 6 to both terms? By subtracting 6 from both terms? 


Answer the same question for =. 
33. Find the value of the fraction eat 
2+ 2 

values 1,10, 100, 1000. (6) When z has the values .1, .01, .001. 


Arrange the results in order of magnitude in each case. 





(a) when « has the 


38. Multiplication and division. The rules for multiplica- 
tion and division are given in Principles V and VI (Art. 35). 


x? — 2ar +a? rita 





Example 1. Find 


g2 pak a2 x? ea a2. 
SOLUTION: at aaa Pon) tae 
eo a? +a? 6? +a? 
LA -Ha 
ad x? —2ax +a? x-a 
Example 2. Divide PY oe savas 


R en 20k a a Peers e+a +a? 
rs x? — a? va? +a? er xr—a xz+a 


Art. 38 ] SIMPLIFICATION OF FRACTIONS 47 


EXERCISES 


Perform the following -multiplications and divisions and 
simplify the results : 
































b Sab a \? 
ti {eg 10. ae, + (oe) : 
7 6ab Sax 
2. 8-57 Leann tt he 
, &%  % ote) 
3. 83 12. wot + 
my EON. ee 
4. 8k - 12. 13. G+) +a 
Sax say. a b 
EeGhy dbx i b—a 
ye ieee eee 
ior. ol " g? ay a? +ab 
ee 40 | ab ab 
ts, a oA iJ 5) 
we m+-1l m—I1 
a to+! m (142) + (242) 
5 n 
19 (a — aes x? — xy + y? 
"a+ y a? -Qar+ 2° 
20 222 wae aie Ig es — 18x + 42. 
ey Ee xe? + Qa 


m2 + 2mn i m? — 4n? 
m?+4n? mn — 2n? 


oe 2 
22. (5 -£41)+(G+241): 

a a a a 

4; et a \2 b 
a a ies) «(1 -i 


48 : FRACTIONS : [Crap ay: 





2x pees uy 
2 tn 2 A ——————————— aa : 
: t vend 2D eee 
- Sipe et me 2 
25. ( - a= a “ys (a — b)?. 
eats 


3 AY es —, ate 
26. (a? + 3a? — 6a — 8) = Aor 


An fh ee ey: 


ee pee ret pyc. — y 

















28. ay gt yi” 76 — 8 

ay arma — a*bsa m3 —bex3 
(m — b?x)8 abm? + 2ab’ma + ab>x? 

30, 2° — 627 + 362 ie 24+ 2162 


zg? — 49 gis 2 A> 


39. Complex fractions. A complex fraction is a:fraction that 
contains a fraction either in its numerator, or in its denomina- 
tor, or in both. Complex fractions are e simplified by the rules 
for division of fractions. 

One of the most useful methods of simplifying complex frac- 
tions consists in multiplying both numerator and denominator 
of the complex fraction by the L.C.M. of the denominators of 
the simple fractions gs make up the terms. 








Example 1. aoe 12(2 - 5) 
* Weie Awe 8 —6y 2708 
IN: 5y 38 aes 
4 12 4D 

mame bac erway 
en a ab(a? — b)| = — - — 
a a-b —b 





S a(a? — b?) + ab(a — b) 

~ b(a? — b?) — ab(a +d) 

_ a(a? — 2b? + ab). 
b2(a + b) 





ArT. 39] 


COMPLEX FRACTIONS 


EXERCISES 


Simplify the following fractions: 








‘tema 
; 913 

a ee 

6 

2-3 
_ 1+4 
peel 
b 

3. i: 
Pe. 
a 


he 


13. 


35 
50.0), 























aa 
be \ 





ara ee" * zr) 
a® + ab + 0° te 


1+(“5+ —) 





49 


c-y +3 
aay z 
J i; 
Seeman 
4 4 
sree eee 
3 
fh dei 
hy 
b + As 
d+- 
i 


00 : FRACTIONS [Cuap. IV. 








ty 
14. ¢+1+ 
ee ttl 
te oa 
16. sree ; 
m—-nt+ 
m+t+tnt 
m—n 


+= +> 
OG, Moe E 2 [See 2 




















DOS. ab abe oe ae 
eaieed ah 
1 1 
aly & ae 
{Lea onl AS Dia os) Pim eb 
1a ae 2 
b+ ; 
d + 
Wea, 
h 
ae: 
Che e = 
12 a 
Le ae 
1 
20. Po 
1, 6 
6h To 
1 
i ee 
: tr t+1rz 


40. Equations involving fractions. In solving an equation 
that involves fractions, it 1s usually convenient to clear of 
fractions by multiplying each member by the L.C.D. of the 
fractions. 


Art. 40] EQUATIONS INVOLVING FRACTIONS ol 


If the unknown occurs in any denominator, multiplying by 
the L.C.D. may or may not introduce new roots that do not 
satisfy the equation to be solved. 














Example 1. Solve = 2 ae Z ire 2. 

SOLUTION : p : Teite: a =e. (1) 
(1) - @& — 1)(@ — 8) gives 5(@ — 3) + 8(@ — 1) = 2(4 — 1) (a — 8). (2) 
Simplifying (2), 2x? — 16x + 24 = 0. (3) 

2(x — 6)(a — 2) =0. (4) 

Hence, =D, 

or Gem 2. 


The roots of (2) are 6 and 2, and both satisfy (1). 


It is important to note that if we should multiply (1) by a common 
denominator other than the L.C.D., roots may be introduced which do not 
satisfy (1). To illustrate, if the members of (1) are multiplied by 
(« — 1)(a — 3)(a — 4), the resulting equation, 


5(@ — 3) (a — 4) + 38(@ — 1) (@ — 4) = 2(e -— I) (x -— 3) (@ — 4), 
has the root x = 4 that does not satisfy (1). 








x —1 
Example 2. Solve TORO eee L 
x -l 
SOLUTION : Ss eae (1) 
(1) - (2? — 1) gives ¢ -l=77 -1, (2) 
or re greene eae UP 
Hence, gi=Oor x = 1. (3) 


The roots of (2) areO and 1. Now = 0 satisfies (1), but x = 1 does not 
satisfy (1) since the left hand member has no meaning when x = 1. Hence, 
the root « = 1 is introduced in clearing of fractions. 

The introduction of roots in clearing of fractions when an unknown 
occurs in the denominator should make clear the importance of checkmg 
each solution by substitution in the equation to be solved. 


52 ’ FRACTIONS [ Cuap. IV. 


EXERCISES 


Solve the following equations and check the results: 


1. 5 Pay 











iy 
4, 2(7x — 10) — $(50 — x) = 20. 
C—O 
DR 1 age ay 

3 I 4 
a AUR OT CBT: 
o(22? +3) fp — 5 
274 +1 27 — 5 
5 4 8 1 
2-6" 2 Qin Aer 1G) 
x—2 x—l 
De 1. BU Be 
x—3 1 4x% —1 
ae cai tae a ae 

21 10 11 
11° > 9 een 
ob Ll eae 
2x+3 47+5 











= 5r — 6. 








Or 





























15. 


— Art. 40] EXERCISES AND PROBLEMS o3 




















16. a—-=c 
x x 
caesar ade area 
ig, Mt _ min 
ie I > 7 
ee eee 
Pp q r 
We phe Uk 
Rika br ——i(i«i‘ié‘iak 
N-x p—-x mp — 2z) 
a1. ——~ 4 = = —O 
m+txrm-z2 nt — 7" 
x 
Sat ars aril Ain od a 
x—.5 on 
oe leh 9.5. 
ier Atel. Oe BLS Ro 
Be Sek SO | 1,00 
2 8 7 


Ban ie a oe 


EXERCISES AND PROBLEMS 


1. What part of a piece of work can A do in 1 day if he can 
do all of the work (a) in 5 days ; (6) in 17 days ; (c) in 33 days ; 
(d) in 52 days; (e) in m days; (f) in m+n days? Tell in 
each case what part he can do in 3 days ; ind days. 

2. A can do a piece of work in m days, and B can do the 
work in n days. What part can each do in 1 day? What 
part can both together do in 1 day? How many days will be 
required by A and B working together to do the work ? 

3. The marked price of a bill of goods is $600. It is dis- 
counted 7%. Find the discount and the selling price. 

4. Find the discount and the selling price if the marked 
price is m dollars and the rate of discount 7%. 


o4 FRACTIONS [Cuap. IV. 


5. Find the marked price and selling price if the discount 
is $40 and the rate of discount 2 %. 

6. Find the marked price and selling price if the discount 
is d dollars and the rate of discount r%. 

7. Find the marked price and the discount if the selling 
price is s dollars and the rate of discount r %. 

8. In a division the dividend, divisor, quotient, and re- 
mainder are represented by D, d, q, and 7, respectively. Express 
D in terms of d, g, andr. Express g in terms of D, d, and r. 

9. If a oranges cost c cents, what is the cost of b oranges? — 
10. What is the average selling price for a horse if m 
horses are sold at a dollars each, and n horses are sold at y 

dollars each ? 

11. What is the average selling price if a articles are sold 
at p cents each, b at gq cents each, and c at r cents each ? 

12. If a train goes m miles in x minutes, how far will it go 
in 1 hour? 

13. At c cents a pound how many pounds of butter can be 
bought for d dollars? 

14. AtS 
bought for $7? For a dollars? For b dollars? 

15. What is the gain per cent on the cost if an article is 
bought for $120 and sold for $150? If bought for x dollars and 
sold for y dollars ? 

16. What number added: to both the numerator and the 
denominator of ;°; gives a fraction equal to 2? 

17. What number added to both the numerator and the 
a 
| 5 

18. If for the sum of the fractions 3 and % we take the sum 
of the numerators divided by the sum of the denominators, 
by how much does the result differ from the true result ? Answer 
“and © 


b d 


dollars per yard how many yards of muslin can be 


denominator of = gives a result equal to=? 


the same question for the fractions 


f 
"Arr. 40] EXERCISES AND PROBLEMS 55 


19. Divide the number n into two parts such that the first 

. part equals 2 of the second. 
20. State the rules of arithmetic for multiplying and dividing 
a fraction by an integer. Apply these rules to multiplying 


and dividing : by c. 


REVIEW EXERCISES AND PROBLEMS ON CHAPTERS I-IV 


Insert expressions in the parentheses so as to form identities. 


1 oct+y—-z=27+( Te 
2.2-y—-z=2x — ( : 

8. 2-y+2=2 — ( ,: 

4,.2+ 6y —6 =2 — 6( Je 

5. a — 25) + 10 =a — 5( a4 

6. ax — ay + bx — by = (a + b)( e 


7. How do you check the solution of an equation ? 


Solve the following for x and check the solutions: 


S42 — Liz = 0.3. 
10. L7 (2 — 2) —0.3(2z% +1) = 08. 











5(2—2) 2(% —4) © 
11. eae 14... = 0. 
or wae 4+ 5 
12: oTal 9 = 22. 
1 2 
ee et oe a 
14. Solve x — (= a a +2) ee +S. (DarTMouTH *) 


15. Remove the parentheses from 
3a — {3a — [3a — (8a — 3a—- 3a) — 3a] — 3a} — 38a, 
and simplify the result. (DARTMOt TH) 
16. Find the value of — 
a — {5b — [a — (8c — 3b) + 2c — 38(a — 2b — o)J}, 
where a = —3, b = 4,c = —5. (Y ALE) 


* Institution that gave the question in an entrance examination. 


56 | FRACTIONS [Cuap. IV. 


17. Find the product: (1 —a#)(14+2)(1 +27)(1 + 24)(1 + 7)(1 4+ 2"). 
(ILLINOIS) 


18. Find the greatest common factor and the least common multiple 
of the three expressions : 








at + a2? + x1, a? + ax + 2, a? — ax +2. (HARVARD) 
a? + b? 
. a 84+ 063 
19. Simplify 1 i preety (YALE) 
GLI6 
20. Divide 62°" — 2522" + 272" — 5 by 2a” — 5. (ILLINOIS) 


Express the following in as few terms as possible : 

21. 4.827? — 2x? 4+ 3.5y? — 2.42y — by? + 2a? — (6.42y — y?). 

22. 6.40? — [1.672 + (7a — 1 + 1.42?) — 3 4 62]. 

23. A person who possesses $15,000 employs part of the money in build- 
ing a house. He invests one third of the money that remains at 6 per cent, 
and the other two thirds at 9 per cent, and from these two investments 


he obtains an annual income of $500. What is the cost of the house? 
(Mass. INstiTuTE) 


24. The admission to an entertainment is 50 cents for adults and 35 


cents for children. If the proceeds from the sale of 100 tickets amounted- 


to $39.50, how many tickets of each kind were sold ? (ILLINOIS) 


25. Divide x increased by 3 by 3x decreased by 4. Divide the result 
by 32? less the binomial 32 — 7, and call the result A. Form the same 
expression where x — 1 takes the place of x and call it B. Divide A by B 
and reduce to a simple fraction. (PRINCETON) 


Factor the following : 


26. 9at — 16. 33. x? — 1. 
27. Yat — 4. 34. xt +1. 

28. (cx —y)? — (x + y)?. 35. 78? — FF 
29. x + 4x2 + 4, 36. x? + 6x? + lla +6. 
30. 35p? + 4pq — 15q?. . 37. xz? — 62? + Ile — 6. 
31. x4 — 162? + 64. 38. m® — né, 


32. x(x +1)(@ +3) +27 +4¢ 43. 39. m® — n8, 


ae 


Art. 40] REVIEW EXERCISES AND PROBLEMS 57 


ord 


Find the values of expressions in Exercises 40-43 for a = 5, x = 8, y = 7, 
in the shortest way possible. 
zy +- a2 x? sae a? 
40. x2 2! a2 xz? + a2 
x? + a? yz? bee a2 


go? —aq* z* 4 a? 


(= ne a" i. v) (5 ean u) 
41. y cia ae BD 


























tee Le 
y x 
ip) of Ur u 
* g(a —1)(x —2)(@ —3) x(x +:1)(z@ — 1) (@ — 2) 
43. : : 


x@+1)(@+2)(@+3) (+I) +2)(@+3)(@ +4) 
44. Reduce to a single fraction 
1 7 r+3 














eee oa A (OHIO STATE) 
ad ea 
45. Simplify : = a ; : - (1 cal =). (Onto Srarn) 


46. Find in the shortest way the numerical value of 
(x? — zy + y”) (x + y), 
if x = 24 and y = 12, 


47. A train running 30 miles an hour requires 21 minutes longer to go 
a certain distance than does a train running 36 miles an hour. How great 
is the distance ? (CORNELL) 

48. Factor, and find the highest common factor and the lowest com- 
mon multiple of the expressions 2x4 — 2° — a#?, 277 + 4% — 3,23 —a2? —x +1. 

49. Simplify : ae - ay - ay (OHIO STATE) 

50. A bicyclist averaging 12 miles per hour is a half mile ahead of an 
automobile running 40 miles per hour. How many minutes before tke 
automobile dashes past him? (STANFORD) 

51. Given the equation 5x — 10 = 3x —2. What extraneous root is 
introduced into the equation by multiplying each member by x — 3? 

52. Verify that 2 is a root of 3r(x — 2) =z? — 4. Is 2 a root of the 
equation obtained by dividing each member of the given equation by 
xz —2? 


58 | FRACTIONS (Crap. IV. 


53. The difference of the squares of two consecutive integers is 41. 
What are the integers ? 

54. Concrete is to be made of a mixture of cement, sand, and gravel. 
The weights of sand and gravel used are equal, but the weight of the cement 
used is $ that of the sand. How many pounds of each are required to make 
7800 pounds of concrete ? 

55. The formula for converting a temperature reading of F degrees 
Fahrenheit into its equivalent temperature of C degrees Centigrade is 


C = $(F —.32). 
Express F in terms of C, and compute F for C = 25 and for C = 42. 


56. The numerator of a fraction is 7 less than the denominator ; if 4 
is subtracted from the numerator and 1 is added to the denominator, the 
resulting fraction equals }. Find the fraction. 





67. Simplify the fraction ! 
a+ Beyrartel (NEBRASKA) 
1+ 
3 =a 


58. A train runs 100 miles in a certain time. If it were to run 5 miles 
an hour slower it would run 20 miles less in the same time. What is the 
rate at which the train runs? (Missouri) 


59. Divide 54 into two parts such that twice the smaller shall exceed 
29 as much as 143 exceeds four times the greater. (WISCONSIN) 
60. Solve the equation 
x+b 2a oa 2e 2a —b 
es eee ) 








(HARVARD) 





oa 
b-—a 
its lowest terms. (HARVARD) 

62. Which takes the lesser force to lift a weight of 100 pounds with a 
crowbar 6 feet long ; (a) when the weight is at the end of the bar and the 
fulerum one foot from the end ; (b) when the fulcrum is at the end of the 
bar and the weight one foot from the end. Estimate your answer, then 
calculate the actual difference. 

63. The distance from San Francisco to Los Angeles is 475 miles. A 
train running 26 miles per hour leaves San Francisco for Los Angeles at 
the same time that a train running 31 miles per hour leaves Los Angeles 
for San Francisco. In how many hours will they meet? (Caurrornia) 


61. Reduce the expression eae (2 = =) to a single fraction in 


CHAPTER V 
FUNCTIONS 


41. Constants and variables. A constant is a symbol which 
represents the same number throughout a discussion. 
Thus, in the formula for the volume V of a cylinder of radius 


rand height h, 
Vc= treks 


the symbol 7 is a constant, whatever values r and h may have. 
In problems dealing with falling bodies the space s through 
which a body falls in time ¢ is given by the formula 


oie agt’, 
_ where g is a constant. | 

The Arabic numerals are, of course, the most common cx- 
amples of constants. 

A variable is a symbol which may represent diiferent num- 
bers in the discussion or problem into which it enters. 
' Thus, in the foregoing illustrations, V, r, hk, s, t are variables. 
Many mathematical expressions contain both variables and con- 
stants. Except in certain geometrical and physical formulas 
it is customary to use the letters a, b, c --- from the beginning 
of the alphabet to denote constants 
and the letters --- x, y, 2 at the end of 
the alphabet, to denote variables. 

Exercise. In Fig. 4, let the point P move 
around the circle, while the points P:, O, and 


Q are fixed. What lines in the figure represent 
variables and what represent constants ? Fig. 4 





42. Definition of a function. Many discussions and prob- 
lems of algebra involve two variables which are so related that 


60 7 ‘FUNCTIONS [Cuap. V. - 


when a value of one is given, a corresponding value of the other 
can be found. Throughout the course in algebra, we have 
had many problems in which two or more such related varia- 
bles occur. In the equation 2” — 3y = 4, if a value be given 
to x, the corresponding value of y can be found. ‘Thus, if 
x=0, y = —$; ifz =1,.y = —4, and so on. 

In evaluating the expression 2?+3x—-—1, we find that 
x? + 3x2 —1= —1 when x=0, 27+ 32% —1=3 when @ =1, 
and so on. Fixing the value of x in the first illustration fixes 
the value of y; in the second illustration fixing the value of x 
fixes the value of x? + 3a — 1. 

DerinitTion. If two variables are so related that when a 
value of one is given, a corresponding value of the other is 
determined, the second variable is called a function of the 
first. | | 

Thus, in the equation 2x — 3y = 4, y is a function of xz. The 
expression x? + 32 — 1, and in general any expression contain- 
ing z, is a function of x We may therefore and shall speak 
of a ‘“‘function of x,” instead of an “ expression involving the 
variable 2.” 


43. Functional notation. If we refer to the same function 
of x a number of times in a discussion, it is convenient to have 
some suggestive abbreviation to represent it. It has become 
customary to represent a function of x by the symbol f(z) 
which is read “f function of z.”’ If another function occurs in 
the same problem it can be represented by F(a), which is then 
read ‘“F major function of 2,”’ while the f(x) is read “f minor 
function of x.” } 

If in a discussion f(x) is used as an abbreviation for the ex- 
pression a* + 323 + 22 — 3, then f(a) means the result when 
a replaces z. Thus, : 


f(a) =a* + 3a? + 2a — 38, 
f(h) =h* + 3h3 + 2h — 8, 
and f(2) = 2443-23? 42-2-3. 


_ Arts. 43, 44] EVALUATION OF FUNCTIONS 61 


meet F(x) = re 
then F(a) = eres 
PQ) =" 

and f(2) = sped 


These illustrations bring out an important point in this 
notation, viz.: If the x in the parentheses of the symbol of f(z) 
is replaced by any other number, the new symbol means the 
value of f(z) when the number is put in place of z. 


44, Evaluation of functions. To evaluate a function of z is 
to put a given number in place of x and to calculate the corre- 
sponding value of the function. 


Thus, if f(z) = 24 + 323 + 27 — 3, we evaluate it for x = 2 by replacing 
x by 2 and reducing. Thus, 
f(@2) =2° +3 -2? 42-2 -—3 =41. 
For xz = 3, we find f(3) = 34+ 3 - 33 +2-3 -—3 = 165. 


EXERCISES 
If f(x) = 2a — 3, find f(1), f), f2), ((-D. 
If f(x) = 3x + 2, find f(0), f(4), f(-4), f(@). 
If f(x) = «?, find fO), f(2), FG), fm). 
If f(x) = 20° — 1, find (4), £7), FAD), fp). 
If f(®) = 16¢?, find f(1), f(2), f(x), f(10). 
If fy) =y +y +1, find f(2), f2.3), fLOD), f@). 


7. If f(s) = <=, find (03), £9), (1.1), FO. 


2 Se eee 


8. It fin) = +" *-, find f(1), $4), $@), SO. 


62 FUNCTIONS [CuHap. V. 


9, If f(x) = 42-42 +3 and F(x) = 2? — 1, find f(2) and 

F(2) ; f(6) and F(6). : 

10. If f(x) = a3 +2 and F(a) = 2x? — 4a — 5, find the quo- 
fC) eZee 

F(1) F(3) 

45. Formulas frequently used. Most of the formulas of 
mensuration learned in arithmetic and geometry 
express variables as functions of other variables. 
Thus, for the area A of a circle of radius r, we 


have the formula 
A= re. 


tients and 


Fic. 5 which expresses the area as a function of the 
radius, so that for a given r the area A can be found. 
The following list of common formulas will be found useful 
for reference. 
1. Area A of a rectangle 
of sides a and b. 
A = ab. 
2. Area A of a parallelo- 
gram of base b and altitude h. b 
A = bh. (Fig. 6.) Fig.6 
3. Area A of a triangle of base b and altitude h. 
A = 3bh. (Fig. 7.) 
4. Area A of a triangle in terms of its sides, 
a, 6 and c. 
A = ~Vs(s —a)(s — b)(s — oc), 
at+bte 
2 





where s = (Fig. 8.) 


Fie. 7 

5. Area A of a circle of radius r and 
‘circumference C. 

A = 4Cr.. 

6. Circumference C of a circle of dia- 
meter d, or of radius r. 

C=7d,orC =2rr. w =3.1416. 
A useful value for practical purposes is 


n= 3.14 a Fic. 8 


Art. 45] USEFUL FORMULAS 63 
7. Area A of a circle of diameter j be 

d or radius r. 
Aj=47d* or A = zr’. 


8. Area A of a trapezoid of | 
bases 6; and bb. and altitude A. | 
(Fig. 9.) bi 





A =4(b: + by)h. Fic. 9 
9. Area A of a regular hexagon of sidea. (Fig. 10.) 
A= BV 3 a2 
2 
a 
Fig. 10 
10. Length ¢ of the hypotenuse of a ; es 
right triangle of sides a and b. (Fig. 11.) 
c = Va? +b. 
b 
Fig. 11 


11. Volume V of a cube of edgea. (Fig. 12.) 
V =a’, 


12. Volume V of a rectangular solid of length 
l, width w and altitude h. (Fig. 13.) 


V =lwh. 





64 FUNCTIONS [Cuap. V. 


13. Volume V of a cylinder of altitude A and radius of 
base r. (Fig. 14.) 


V = rh. 


14. Volume V of a pyramid of altitude h and area of base 
B. (Fig. 15.) 





V =4Bh. 


15. Volume V of a cone of altitude A and radius 
of base r. (Fig. 16.) 


V= $777k. 


16. Surface S of a sphere of radius 7, or dia- 
meter d. 
S =47r?, orS = 1d?. Fig. 15 


17. Volume V of a sphere of radius r, or dia- 


meter d. 
V = g7r,or V = 47d. 


18. Volume V of a spherical segment (a slice: of 
a sphere between two parallel cutting planes), where 
his the altitude and a and } the radu of the two 
bases. (Fig. 17.) Fig. 16 


19. Surface S of a zone (the portion of the 
surface of a sphere lying between two parallel 
cutting planes), where A is the distance between 


the cutting planes and r the radius of the sphere. eas 


S = 2arh. Fig. 17 


‘Arr. 45] EXERCISES AND PROBLEMS 65 


EXERCISES AND PROBLEMS 


1. The length of a rectangle isa. The length is three times 
the height. Express the area as a function of a. 

_ 2. The altitude of a triangle is 2 inches less than the base. 
Express the area in terms of the base. 

3. The sum of the two bases of a trapezoid equals three times 
the altitude h. Express the area as a function of the altitude. 

4. Iixpress the area of a circle as a function of the circum- 
ference c. 

5. The sides of a rectangular solid are a, a+ 2, a +3. 
Express the volume as a function of a. 

6. The lengths of the sides of a rectangular solid are con- 
secutive integers. The shorter side is of length J. Let V be 
the volume, then V = f(l). Find f(J). 

7. The base of a pyramid is a square whose side equals the 
altitude h. Express the volume of the pyramid as a function of 
the altitude. 

8. Find the volume of a pyramid of altitude 10 and square 
base of side 4. 

9. Find the volume of a cone of altitude 7 and radius of 
base 4. 

10. The altitude of a cone is one greater than three times 
the radius r of the base. Express the volume as a function of 
the radius. 

11. Express the Ae of a sphere in terms of the circum- 
ference C of a great circle (the largest circle that can be drawn 
om the sphere). 

12. Let V be the volume of a sphere, and c the circumference 
wf a great circle. Then V =f(c). Find f(c). 

13. Find the volume of a spherical shell. The thickness of 
she shell is one inch and outside diameter 10 inches. 

14. Express the following sentence in functional symbols. 
[he number N of eggs bought for a given amount of money 
lepends upon the price c per dozen. 


66 FUNCTIONS [Cap V. 


15. Express in functional symbols: If the age x of the 
applicant for life insurance is known, the premium P is fixed. 

16. If d is the distance passed over in ¢ hours by an aero- 
plane going 70 miles an hour, then d = f(t). Find f(é). 


46. System of codrdinates. Let X’X and Y’Y (Fig. 18) 
be two straight lines meeting at right angles. Let them be 
considered as two number scales with the point of intersection 
as the zero point of each. Let P be any point in the plane. 
From it drop perpendiculars to the two lines. Let x represent 
the perpendicular to Y’Y and y the perpendicular to X’X. 

| If P lies to the left of Y’Y, x is 
to be considered negative. If 
P lies below X’X, then y is 
negative. It is clear that no 
matter where P is in the plane, 
there corresponds to it one and 
only one pair of perpendiculars, 
x and y. 

The lines of reference X’X 
and Y’Y are called the coérdi- 
nate axes, and their intersec- 
tion is called the origin. The 
first line is called the X-axis, and the second the Y-axis. The 
perpendicular to the X-axis from a given point in the plane is 
called the ordinate or y value of the point. The perpendicular 
to the Y-axis is called the abscissa or x value of the point. i 

If we have two numbers given we can find one and only one 
point P which l:as the first number for its abscissa and the 
second for its orcinate. If, for example, the numbers are 27 
and —5, we measure from the origin, in the positive direction 
a distance 2 on the X-axis and at this point we erect a perpen- 
dicular and measure downwards a distance 5. We have then 
located a point whose x is 2 and whose y is —5. This point may : 
be represented by the symbol (2,—5). The symbol (a, 6b) 





Art. 46] PLOTTING OF POINTS . 67 


denotes a point whose abscissa is a and whose ordinate is b. 
: The symbol P(a, b) is sometimes used and is read, “the point 

whose coordinates are a and b.”’ 
_ When a point is located in the manner described above, it 
is said to be plotted. In plotting points and obtaining the 
geometrical pictures we are about to make, it will be convenient 
to use codrdinate paper (Fig. 19). Then the side of a square 
may be taken as the unit of length to represent a number. 

















To plot a point, count off from the origin along the X-axis 
the number of divisions required to represent the abscissa 
and from the point thus determined count off the number of 
divisions parallel to the Y-axis required to represent the 
ordinate. 


EXERCISES AND PROBLEMS 


1. Plot the points (2, 4), (—2, 4), (2, —4), (—2, —A4). 

2. Plot the points (8, 7), (7, 3), (0, 0), (0, 1), (2, 0). 

3. Plot the points bad by, ook, 0), (3, 3) (3, an Ge eit 
(-3, —$), (— 3, 0). 

4. Draw the triangle whose vertices are (0, 0), (0, 1), (2, 0). 

5. Draw the triangle whose vertices are (1, 3), (—1, —1), 
(2, —2). 


68 FUNCTIONS [Cuap. V. 


6. Draw the quadrilateral whose vertices are (0, 0), (2, 3), 
(5, —1), (3, —3). 

7. A square of side 2 has one corner at the point (2,1). If 
the sides of the square are parallel to the codrdinate axes, what 
are the codrdinates of the points that may be at the other cor- 
ners of the square ? 

8. A line is bisected by the origin. One end of the line is 
the point (2, —5). What are the codrdinates of the other end ? 

9. A circle of radius 2 with its center at the origin intersects 
the codrdinate axes at what points? At what points does it — 
intersect the straight line which passes through the points 
(2, 2), (—2,. —2) : 

10. Let the X-axis be considered as an east and west line, 
and the Y-axis as a north and south line. The following 
points on a river indicate its general course. Map the river 
from x«-= —4 to 2=+4. (—4, —22); 
(— 33, es) La 2} (27; =a) 
2 14), Looe at) a i (—3, 3); 
(0,1), (2,8), (1, 18), G, 8), 2, 1), a, 9), 
(3, uy (33, ak a 135 

47%. Graph of a function. The change 
in a function can be represented to the eye 
on codrdinate paper. For example, the 
change in the area of a square due to the 
change in the length of the side can be 
represented to the eye as follows: Let A 
be the area and a the length of the side. 
Construct a table showing the area for 
—-X various values of a. 














Peed eena., on tee a=-|3/1\| 3/2) 
Fig. 20 A=|1|1|2;|4 7 


Draw coérdinate axes on paper and plot the points (4, 4), (1, 1), 
(2, 21), and so on (Fig. 20). Connect the points by a smooth — 
curve. 


Art. 47 | GRAPH OF A FUNCTION 69 


The fact that the area increases more rapidly than the side 


_is shown in the upward bending of the curve. 


By this method, which is the same as that used to map a river 


) (Exercise 10, Art. 46), any function may be represented on co- 
ordinate paper. This representation of a function is called the 


graph of the function. The graph of f(x) contains all the points 
whose coérdinates are [x, f(x) ] and no other points. 


Example. Obtain the graph of 5 — 1 for values between —5 and +5. 


Let f(x) = ; —1. The object is to present a picture which will exhibit 


the values of f(x) which correspond to different values of x. Any value of x 











with the corresponding value of f(x) determines a point whose ordinate is 
f(x). Assuming values of x and computing the corresponding values of 
f(x), we obtain the following table. 





emmy 2 | 23 | 4 | 5| -1| -14| -2 | -3.| -4 | 5 
f(z) = —1 |)-# 10] 414111281 -$| -12 1 -2 | -3 | -3 | -33 
The corresponding points (0, —1), (1, —2),---, are plotted in Fig. 21. 


It is seen from the figure that all the points lie on a straight line. This 


shows that the function 5 1 increases uniformly as x increases. 


70 | FUNCTIONS [Cuap. V. 


EXERCISES AND PROBLEMS 


1. Draw a curve showing the change in the volume of a cube 
as the length of the edge x changes from 0 to 5. 

2. One side of a rectangle is x, the other side is x + 1. 
Show by a curve the change in the area as 2 changes from 0 
to 10. 

3. Show by a graph the change of the circumference as the 
radius 7 changes. 

4. Show by a graph the change in the area of a circle as the 
diameter d changes. 

5. Exhibit graphically the change in the volume of a sphere 
as the diameter d changes. 

Graph the following functions on codrdinate paper from 
x = —4tox = 4, plotting at least 17 points. 


6. «+4 2. TO 
7 2¢+1. 11,.-2* a 
8. 27 — 1. 12. 2742. 
rye ae 13. 27 +241. 
3 14. l+2—-— 27. 
15. 1 —2 — x. 


MISCELLANEOUS EXERCISES AND PROBLEMS 
Evaluate the following functions : 


Late) _ at fore =1,2=3,2% = = =e 


2. f(s) = s* — 7 fors = —4,8 =1.2,3 =17, 
. f(m) = m? + 38m +1, for m = —4, m = 13, m = —10. 


is) 


4. If f(a) = 0 2a? + a, find fQ), Te f(0). 
6. Tse) = SEER, find f08), f(-3), S@), f0). 


| Arr. 47] EXERCISES AND PROBLEMS 71 


Beate y(n) os 4° find f(2*), f(a + 1). 


m7, If f(r) = a — 1, and F(x) = 1 — 2x, find F(2) + f(2). 

8. The dimensions of a rectangular solid are consecutive 

integers. The shortest edge is represented by x. Express the 
volume. Find the volume when x = 7; when z = 11. 


| 9. Arrange the following numbers in order: f(—38), f(—2), 
ie (— 1); f(0), £2), F@), F(22), f(3), where fv) = 2? +2 +1. 








10. Fill out the table where f(a) = ue ai arch 
a7 +1 
a=|-3|-2/-1/0|1|[2|3|[41|5 





f(a) =| | | ge ae et | 
11. Fill out the table for f(x) = 10x + 1002?. 
Mereroyer | 203 | 41.5 | 6] .7 ] 8] 9] 1 
RS a Rea cain ce es | 
Peeeutihe table for f(y) - 5 























On codrdinate paper graph the following functions from 
x= —d tox = Dd: 








x x—l x+1 
13: = ise 17, ZH. 
14. : bat? 16. x — 22. 18. (x +1)(x +2). 


19. If f(x) = 2* — 223 — 27? + 2x show that: f(2) =f(1) = 
Bie b) = f(0). 

20. For the function in Exercise 19, show that f(3) = f(—2). 

21. If f(a) = a? + 1, find f(a + 1). 

Peeeiiaieer + 7 4 1, find f(2x — 1). 

23. If f(a) = 2a + 1, find f(2z + 1). 


72 FUNCTIONS [Cuap. V. 
24, If f(x) = 22 —2x —1, find i(;). 


25. It F(a ee find (3): 


ate 





26, Lo yia)= 1’ find f(# — 2). 


Bie ee Hrontts the length of an adult’s foot in inches 
and s the size of the shoe, then 
s 
F = 3 + 8. 
Represent by a diagram the relation between length of foot 
and size of shoe for sizes 1 to 12. 

28. For the 14 days preceding the opening of the European 
war the daily average price of 20 well-known stocks was as 
follows: 100.70, 100.63, 100.01, 104.49, 98.30, 98.77, 98.49, 
97.95, 97.05, 97.16, 96.58, 93.14, 94.12, 89.41. Show this de- 
cline in price by a diagram. 

Hint: To save work in handling these large numbers, 80 may be sub- 
tracted from each of the above numbers and the differences plotted. 


29. The weight of one mile of fine iron wire depends on the 
diameter of the wire according to the table: 


Diameter in inches.|.007|.008|.009|.010|.011|.012|.013|.014|.015|.016|.017|.018]|.019|.020 
Weight in pounds. |.7 |.9 |1.1|1.4|1.7|2.0|2.4 | 2.7| 3.1|3.6|4.0]| 4.5] 5.0 | 5.6 





Exhibit graphically. 


30. The weight of water changes slightly with the tem- 
perature according to the table: 





Temperature Fahrenheit. | 40 | 50 | 60 | 70 | 80 | 90 | 100] 110] 120 
Weight of cubic foot water in lbs. |62.42|62.41162.37|62.31|62.28|62.13]62.02|61.89|61.74 
Temperature Fahrenheit. | 130| 140| 150| 160| 170! 180| 190] 200] 210 








Weight of cubic foot water in lbs. |61.56|61.37|61.18|60.98|60.77|60.55|60.32|60.07|59.82 
Exhibit graphically. 


Hint: The number 60 may be subtracted from each of the above weights 
and the resulting graph will be unchanged in shape. 


Art. 47] PROBLEMS 73 


31. The weight which a manila rope will safely bear de- 


' pends upon the diameter of the rope according to the table: 


- Diameter of rope in inches. 





bie] 1 [1d] 1d1 121 2 | 23 [25122] 3 
1|2| 


| Z| 
Maximum safe load in tons. | 4 | 34 | 54 | 7} [1041 133 | 174] 21 | 25 | 293 


Represent graphically. 


Hint: Let 2 inches be the horizontal unit and one quarter of an inch the 
vertical unit. 


CHAPTER VI 
SYSTEMS OF LINEAR EQUATIONS 


48. Graphs of linear equations. The graphs of some func- 
tions are found in Chapter V. 


REVIEW QUESTIONS 


1. What is meant by (1) codrdinate axes, (2) coérdinates of a point? 

2. Locate points represented 
by the symbols (2, 5), (—2, 5), 
(2, —5), (-2, —5). 

3. What are the approxi- 
mate codrdinates of P,Q, R, S, 
T, U, V, and W in Fig. 22? 

4. What is meant by the 
origin? What are its codérdi- 
nates ? 


An equation of the form 


ax +by+ce=0 (1) 





is called a linear equation. 
When 6 = 0, this equation may be put into the form 
Wine es 


Ba bib (2) 
Since in (2), for any given values of a and b, we may assign 
x and compute corresponding values of y, the equation gives 
any number of pairs of values to plot as codrdinates of 
points. In other words, y is given as a function of a in (2), 
and the graph of this function is called the graph or locus of 
equation (1). 


Arr. 48] GRAPHS OF LINEAR EQUATIONS 75 


When a or 6 is zero, the graph is a line parallel to the 
X-axis or to the Y-axis. Thus, the equation 
j ie =A) 


_ gives a line parallel to the Y-axis, and 2 units to the left of that 
AXIS. 


that the points determined fall on 


The linear equation is so called because its locus is a straight 
line. 3 
EXERCISES AND PROBLEMS 
Draw the loci of the following equations: 
y= 1. 7 


SOLuTIoN: Let x = 0, and we have y = -1. Let y =0, and we have 


Ne 


*=1. The points (0, —1), (1, 0) 
being located, we draw a straight 
line through them. If we are 
given that the graph is a straight 
line, why is it sufficient to plot 
only two points? Assign further 
values, 1, 2, 3, 4, 5 to x and verify 














the locus plotted. xu xX 
2. y — 2x = 4. 
Been oy = 2. 
4. 3x2 — 2y = 6. 
6. 3x + 2y = 6. 
6. 7x + dy = 0. 
7 (x+ dy = 4. | 
8. ies ai= 0. Seale, 5 aici 
Hint: Show the position of all Ys 
points for which y = 2. Pia: 23 
Breer B=: 0), 


Hint: Show the position of all points for which x = 8. 


10. The relation between temperature in degrees Fahrenheit 


F and degrees Centigrade C is given by 


9 
F=,C +32. 


Construct the locus of this equation. 


76 SYSTEMS OF LINEAR EQUATIONS  [Cnap. VI. 


11. The speed v in feet per second of a falling body at time 


t is given by » = 32.2t + 10 
aeys 


where 10 denotes the speed at the instant from which time is 
measured. Construct the locus that shows the relation. 

12. Where does the locus 38x — y = 6 cut the X-axis? The 
Y-axis? Answer the same question for the loci 2% — 4y = 9, 
and x + 3y = 0. 

13. Plot the loci of x —y=5 and of 2x+y=4. Do the two 
lines have any point in common? What are the codrdinates 
of this point? Do these codrdinates satisfy both equations ? 

14. Plot-x.— y = 5 and x —y = 10. Do the two lines have 
a point in common ? 

15. Plot x — y = 5 and 2x — 2y = 10. Do the graphs have 
points in common ? 

49. Graphical solution of a system of linear equations. As 
stated in Art. 48, the locus of any 
linear equation in x and y is a 
straight line. Any such equation 
is satisfied by an indefinitely large 
number of pairs of values of x 
andy; that is, by the codrdi- 
nates of all points on the locus. — 
In the graphical solution of the 
system of two equations, we seek 
the codrdinates of points common 
to the loci of the two equations. 

As the loci are two straight 
| [Naa lines, three cases arise: 

Seale, 2 spaces=1 Unit (1) In general, two lines inter- 
sect in one and only one point. 
This situation is illustrated (Fig. 
24) by the graphical solution of 
Eye, 
y + 2a = 4. 





Yu 
Fic. 24 


- Arr. 49] GRAPHICAL SOLUTIONS (ii 


The solution is x = 2, y = 0. Hence, in general, a system of 
' two linear equations has one and only one solution, but two 
exceptions are now to be noted. 

(2) Two lines may be parallel, and thus have no common 
point. This situation is illustrated (Fig. 25) by the graphs of 
the equations of the system 

xr—y=2, 
2x — 2y = 9. 


~ When the loci are two parallel | 
lines, there is no pair of num- xi. —— | aX 
bers that satisfies both equa- 
tions, and the equations are 


said to be incompatible or 
inconsistent. 


(3) Two lines may be co- 
incident, and thus have an 
indefinitely large number of 
points in common. ‘Thus, if 
we plot « — y = 2, and 3a — 
3y = 6, we shall find that 
the graphs coincide. There 
are an indefinitely large num- 
ber of pairs of values that satisfy this system. The two equa- 
tions of the system are in this case equivalent or dependent. 











Pre 


Scale, 2 spaces=1 Unit 








Pia. 25 


EXERCISES 
Find the solutions (if they exist) of the following equations : 
Piso  ~_— 
hates y=. 19, 4. 3a + 4y = 24, 
26 —y = 1. ox — dy = 11. ‘ 
g: or + 3y = 4, mae 
g+1.5y = 3. y—«x =4. 
3. 2e+y =9, 6. 2x + dy = 10, 


6a — dy = 23. 5a + dy = 7. 


78 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 


7. 7x + 8y = 15, 8. dx + 4y = 6, 
Pe 4flen(, 5a + 4y = 18. 


9. Plot the locus of y + 2% = 5, then multiply both members 
by 3, and plot the locus of the resulting equation. Compare 
the graphs. 

10. Plot the loci of y + Qn = 5, and of 8x — 2y = 4. Then 
form an equation by adding the mentee of these two equa- 
tions and plot the locus of the resulting equation. What are 
the coordinates of the point common to the three loci ? 


50. Elimination. The process of deriving from a system of 
n equations, a new system of n — 1 equations with one fewer 
unknowns, is called elimination. Thus, when the given sys- 
tem consists of two equations in two unknowns, the new system 
consists of a single equation in one unknown. 


51. Elimination by addition or subtraction. Elimination by 
addition or subtraction consists first in multiplying the mem- 
bers of the two equations by such numbers as will make numer- 
ically equal the coefficients of the unknown to be eliminated. 


Then, by addition or subtraction of the members, this oe 


is eliminated. 


Example. Solve 6a — 5y = 14, (1) 
1% + 2y = 32. (2) 
Soutution : Eliminate y as follows : 
(1) - 2* gives 12% — 10y = 28. (3) 
(2) - 5 gives 30x + 10y = 160.. (4) 
(3) + (4) gives © 47x = 188, (5) 
x =4. (6) 
gave 4 for x in (1) gives 
24 — dy = 14, (7) 
—5y = —10, 
y= a (8) 
Cueck: Substituting 4 for x and 2 for y in (1) and (2) gives 
24 —10 = 14, (9) 
28 + 4 = 32. (10) 


~ * (1) - 2 means “ the members of equation (1) multiplied by 2.” 


ek ot ae, Lee eh 2 a 


- a ee, ee ae ee 


Art. 52] ELIMINATION 79 


52. Elimination by substitution. Elimination by substitu- 
tion consists In expressing one unknown in terms of the other 
in one equation, and substituting this result in the other equa- 
tion, thus obtaining one equation in one unknown. 


Example. Solve tz —3y = 26, (1) 

Qn + lly = 43. (2) 

SoLuTion: From (2), 2x = 43 — lly, (3) 
43 — 11 

and C= Teas (4) 


Substituting 8 ty for x in (1) gives 


7 STU _ 3y = 26. (5) 
(5) - 2 gives 7(43 — 1ly) — 6y = 52. (6) 

Then —83y = —249, 
yrs. (7) 
Substituting 3 for y in (4) gives) a = 5. (8) 

Cueck: Substituting 5 for x and 3 for y in (1) and (2) gives 
35 — 9 = 26, (9) 
10 + 33 = 43, (10) 
EXERCISES 


Solve the following systems of equations, and check the 
results : 


1. 6x + 5y = 22 y 
; 6. ks, 
4x —y = 6. gens: 4 
2. « + 2Qy = 14, ae 
2x + 8y = 23. 7. 5 + 8y = 15, 
3. 2a + 5y = 4, y 
4a — 10y = 48. + 4 = 387. 
4, 87+ 2y+5=0, x 
6x + 10y + 16 = 0. 8. 5 + oy +1 =9, 
5. 2(2+y) = 16, x 2 
= + ¥ + 
3(x — y) = 6. eed 


80 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 





9. 0.8¢ + O.ly = 0.19, 11. ax + by = 2ab, 
0.62 + 0.9y = 0.39. bx + ay = a? + 07. 
x+5 
10. —5— + Sy = 9, 12, 2¢+y=m+n, 
mx — ny = m? — n?, 
psy +2=5 
2 
dd. Systems of linear equations solved by determinants. 
Let 
ax + by = a, (1) 
art + boy = &, (2) 


be a system of linear equations in two unknowns. We set the 
problem of finding values of 2 and y that will satisfy both 
equations. Multiply the members of (1) by 62: and those of 
(2) by —b. Adding the members of the resulting equations, 
we obtain 


(abe a= dob) x = boty — Dice, 
>) bocy = bice 


or t= 
abs — dob; 


) provided ibe — deb, ¥ 0. 

In a similar manner, by multiplying (1) by —a: and (2) by 

a, and adding, we obtain 
1C2 — Ae : 
hr ER provided aibe — doh, # 0. 

We note that the denominators of the above fractions are 
alike. This denominator may be conveniently denoted by the 
symbol 
ay by 
2 be 








which is called a determinant. Since it has two rows and two 
columns, it is said to be of the second order. The letters a, 
bi, a2, be, are called the elements of the determinant, and a, 
bs are said to constitute the principal diagonal. A determinant 
of the second order then represents the number which is ob- 


Art. 53] SOLUTIONS BY DETERMINANTS 81 
tained by subtracting from the product of the terms in the 
principal diagonal, the product of the other two terms. Thus, 


1 2 
o 4 


ai Se Gee 








[= ew — vz, | 


Although we have solved systems of linear equations by cer- 
tain methods of elimination (Arts. 51, 52) it is often better 
to employ determinants. The use of determinants introduces 
us to an important instrument that is essential in the study of 
more advanced algebra. 

Using the determinant notation, we may now write the solu- 
tions of our equations in the form 




















C1 by mM A 

C2 be a2 C2 
a olay bil 

dz be az be 














We note that the numerator of the solution for x is obtained 
from the denominator by substituting in place of a, a2, which 
are the coefficients of « in the equations to be solved, the 
known terms G, ¢. In a similar manner, in the numerator 
of the solution for y we replace bi, be by c, @, respectively. 








EXERCISES 

1. Solvex+y = 3, 

22 + dy = 1. 
SOLUTION: 

sya | Leys 

—] Sg | om 
a ae a 
2 3 2 3 





Solve the following pairs of equations, using determinants: 


2. Tx + 9y = 41, 3. +5 = 9, 
oO 


82 SYSTEMS OF LINEAR EQUATIONS L[Cuap. VI. 





4, 3x -—y+1=0. 5. 52 — 2y = 26, 
sy —4% +10=0. © 4x + 3y = 7. 
Ps Iie — dy _ 8a t+y | 
; 22 32 
8x — by = 1. 


Hint: First clear of fractions and simplify. 


4r —3y—% 32 2y 95 


ie 


5 = LD etbaet) 
3 pre PAS 15 6 vel 
8. ax — by = 0, 
Lie 9f =. C. 
Y 
iF 4(z — 1) =3 + 32 — 5 
24 — 3y + 2 = 0. 
10. a(a + y) + D(a — y) = a? + D?, 
a(at+y)+b0(y—2) =a — B. 
11. (m+ n)x — (m — n)y = 21m, 
(m + 1)x — (m — l)y = 2mn. 
Az. 2 — oy — 6, 14. 0.32 + 0.2y = 4, 
Ax — dy = 24. 0.7x — 0.6y = 4. 
13. 37 + 4y =7 — dy, 15. mx + ny = 2mn, 
z—y = 6 — 2z. nz +my =m? + n?. 


54. Determinants of the third order. The square array of 
nine numbers with bars on the sides 


ay bi C1 
(0b) bs C2 
a3 bs C3 








is a convenient abbreviation for the expression 
aibec3 + byc2a3 + Cidebs — asbecr — b3@Q, — C3Q2by, (1) 


and is called a determinant of the third order. As in the case _ 
of the determinant of the second order, the letters a, bi, --- 


Art. 54] SOLUTIONS BY DETERMINANTS 83 


are called the elements, and the letters ai, be, c3, form the prin- 
cipal diagonal. The expression (1) is called the expansion or 
development of the determinant. It is seen that each term of 
the expansion consists of the product of three elements, no 
two of which lie in the same row or in the same column. Any 
determinant of the third order may be easily expanded as 
follows. Rewrite the first and second columns to the right of 
the determinant. From the diagonals 


Qy by 
Aly b2 
a3 bs 


QA bi C1 
dz be & 
a3 bs C3 











running down from left to right we find the first three terms 
of (1). From the diagonals running up from left to right we 
find the last three terms of (1), after prefixing negative signs. 


EXERCISES 
Obtain the expansions of the following determinants: 
1(13 4 
23 
Fra 








When the first and second columns are written to the right 
of the determinant, we have 


13 4)1 3 
+ od Gas aw | 
MesyDi| b 3 


The diagonals running down from left to right are 
Pape oo, 1:4, 2; 3. 
The diagonals running up from left to right are 
eto: 3, 1; 5, 2, 3. 
The expansion then is 
1:7-543-3-144-2-3 —1-7-4-—3-3-1-—5-2-3 =1. 











84 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 






































SVL eS 5.0 ieee 
3.3 12 Om bh 
es hag 0 az be 

Bret earceaeL 6. ld ae 
3 2-4 Q) ey 
See ao u Ok 

4, |-1 1 I 7s lta ene 
alert 20u 2 
1 1-1 6 34 

Sia ay 
Lode 
x4y 








55. Solution of three equations with three unknowns. 
Let the three equations be 


axe + by + a2 = dh, (1) 
dex + bey + @Z = de, (2) 
asx + bsy + c32 = ds. (3) 


Multiplying (1) and (2) by be and —b; respectively and adding, 
we get 


(aibe 2 doby)x + (cibe = biC2)z = bed, ee bid. (4) 
Eliminating y in a similar manner from (1) and (3), we find 
(asby — aybs)a +- (bics _ bsc1) 2 = bid3 — bsdy. (5) 
We now have two equations in two unknowns x and z._ Elimi- | 
nating 2 from these two, we find 
[(aibe — agbi) (bcs — bse1) — (asb1 — aibs) (bec, — dice) |x 
= (dibs _ dob;) (c3by — b3C1) oe (dsbr a dybs) (becy = bice), . 
which after some simplification gives us 


ms diboc3 + debscr + dsbic2 — dibsce — dsb. — dob1¢3. 7 
ayboc3 + debsc: + asbice — arbsc, — azsbec, — aebic 








Art. 55] SOLUTIONS BY DETERMINANTS 85 


The denominator is the development of the determinant in 
Art. 54, while the numerator is the same as the denominator 
with a replaced by d. Hence, we can write the solution for x 
in the form 


dy bi C1 
do bo C2 
ds bs C3 
’ 
an bi Cy 
(lg bo C2 
a3 bs C3 





provided the determinant in the denominator is not zero. 


In a similar way, we can find the value of y and of 2. 





a dy C1 a by dy 
a2 do C2 a2 be dy 
ag ds C3 a3 bs ds 
4 ian bs a) Seariby. ci). 
d2 be C2 2 bs " 
a3 b3 C3 a3 b3 C3 


The denominators in the expressions for 2, y, and 2 are the 
same, while the numerators are obtained from the denomina- 
tors by replacing the coefficients of the unknown in question by 
the known terms. For example, in the numerator of y, the 
knowns dj, de, ds replace 61, bs, bs, respectively. 


Historical note on determinants. It appears that the German philos- 
opher and mathematician Leibnitz first used determinants in solving equa- 
tions. In a letter to a friend, written in 1693, he introduced the notion of 
-determinants in his efforts to simplify expressions arising in the elimina- 
tion of certain unknowns from systems of equations. It seems, however, 
that Leibnitz made very little use of the method and that it did not become 
known to mathematicians ; for, in 1750 Cramer, a professor at Geneva, 
rediscovered the method, and his work is accepted as forming the begin- 
ning of the development of the subject. 
































86 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 
> EXERCISES 
Solve: 
1 #«—y—z= —-6, 
2x+y+2=0, 
ax — dy + 82 = 13. 
SOLUTION : -§6 -1 -l 
0 1 1) 
8.225" sl; 273 ae 
71 =f ip 29saee 
2 1 1 
3 —9d 8 
1 -6 -l 1 -1 -6 
(2 0 1 2 Le 0 
_ (3.418 8' 39, «8, 8 =o ie ae 
Glee. Gel aoa “| [Ii 9 eo 
2 1 1 2 1 1 
3.5 RS 3) Doe Ee 
2. 32 + 2y —2 = 4, 7. it aoe 
ba — 3y + 22 = 5, y 
6x — 4y + 32 = 7. 2+2_. 
3. t+y+2=1, Yee 
ox + 2y + 7z = 1, eatin. 
15a — Ay + 82 = 18. Ty ae 
4, 2x + 3y + 52 = 2, 8. ax + by =a, 
ba — y + 42 = 5, by + cz = b, 
7a — 2y + 62 = 5. ax + cz = 
5. «+ 2y+z2=0, 9. ax + by — cz = 2ab, 


2x +y+ 22 =3, 


Ay — 6y + 82 = 14. 


sete = als 
y+z=2, 
~e+2=4. 


10. 


by — ax + cz = 2be, 
ax — by + cz = 2ac. 


x+y = 3a, 
u.+ 2 = 4a; 
y+z2= 5d. 


Art. 55] EXERCISES AND PROBLEMS 87 


Lele One: 
D eeGurta 
Cae 
Sit See 
ars 
973 13. 


Ton Aero = Or eo, 
10A + 2B — 36C = -1. 


PROBLEMS 


1. The sum of two numbers is 95, and their difference is 
15. Find the numbers. 

2. Thirty-four tons of slate are to be hauled in two-horse 
and four-horse loads. It is found that this will require 9 two- 
horse and 4 four-horse loads. Assuming a four-horse load is 
twice a two-horse load, how much is carried on each kind of 
load ? 

3. A workman is engaged for 30 days on the following terms. 
He is to receive $3 for each day that he works, and is to be 
fined $1.50 for each day he is idle. He received $63 for the 
30 days. How many days did he work ? 


4. Two books cost a cents. The one cost a cents more 


5 
than the other. Find the cost of each book. 
5. A was m times as old as B a years ago, and will be n 
times as old as B in b years from now. Find the ages of each 
. in terms of a, b, m, and n. 


PROBLEMS PERTAINING TO DIGITS 


6. Two numbers are written with the same two digits ; the 
difference of the two numbers is 45 and the sum of the digits 
is 9. What are the numbers? 


88 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 


7. The sum of the two digits with which a number is writ- 
ten is 15. The digit in tens’ place is one larger than that in 
units’ place. Find the number. 


8. There is a number of three digits whose sum is 14. The © 


sum of the digits in tens’ and hundreds’ places is equal to that 
in units’ place. If the digits were written in reverse order, the 
resulting number would be 495 greater than the given number. 
Find the number. 

9. There is a number of three digits whose sum is 12. The 
digits in tens’ and units’ places are equal. That in hundreds’ 
place equals twice that in units’ place. Find the number. 


PROBLEMS PERTAINING TO. MIXTURES 


10. What quantities of silver 72% pure and 84.8% pure 
must be mixed together to give 8 ounces of silver 80 % 
pure ? 

11. What quantities of two liquids, one 95% alcohol and the 
other 15 % alcohol, must be used to give a 10-gallon mixture of 
45 % alcohol ? 

12. The crown of Hiero of Syracuse was part gold and part 
silver. It weighed 20 pounds, and lost 1.25 pounds when 
weighed in water. How much gold and how much silver did 
it contain if 19.25 pounds of gold and 10.5 pounds of silver 
each lose a pound when weighed in water ? 


EXPLANATION OF WEIGHT IN WatTER. A body like a piece of-gold or 
silver when weighed in water loses an amount of weight equal to the 
weight of the water displaced. Thus, if x and y denote the weights of 
gold and silver, 


© y 
(oe 105 24°; 

13. One bar of metal is 20% pure silver and another 
is 12% pure silver. How many ounces of each bar must 
be used, if, when the parts taken are melted together, a bar 
weighing 40 ounces is obtained, of which 15% is pure silver? 


eae - + 


Arr. 55] PROBLEMS 89 


PROBLEMS PERTAINING TO BUSINESS 


14. A man has $35,000 at interest. For one part he receives 
35% and for the other 4%. His income from the money is 
$1300 per year. How is the money divided ? 

15. What is the capital of a person whose income is $1302, 
when he has + of it invested at 8%, $ at 6%, and the remainder 
at 5%? 

16. The garrison of a certain town consists of 240 men who 
receive $3850 as monthly pay. The monthly pay of a cavalry- 
man is $20 and of an infantryman $15. How many men of 
each class are there in the garrison ? 

17. A farm laborer engaged to work 90 days at $1.50 per day 
and board. For days which he was idle, he was to pay $1 for 
board. At the end of the time he received $97.50 net. How 
many days did he work ? 

18. A man distributed a dollars among n persons of two 
classes, the one class receiving b dollars each, and the other c 
dollars each. Find in terms of a, b, and c the number that 
received 6 dollars and the number that received c dollars. 


PROBLEMS PERTAINING TO AVERAGES 


19. Four numbers have the property, that when succes- 
sively the arithmetical average of three of them is added to 
the fourth, the numbers 29, 23, 21, 17 result. What are the 
numbers ? 

20. A student has three grades which give an average 85. 
The sum of two extreme grades exceeds the intermediate grade 
by 75. The highest grade exceeds the intermediate by 3. 
Find the grades. 

21. A high school student received the same grade in solid 
geometry and in second-year algebra, and 3% less in plane 
geometry than in first-year algebra. To find the average grade 
of the student in all his mathematics, the grades in second-year 
algebra and in solid geometry were each multiplied by 3, and 


90 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 


added to the grades in first-year algebra and plane geometry. 
The sum thus obtained was divided by 3. The average thus 
computed equalled 87. If the grades in first and second year 
algebra were interchanged, the average would be decreased by 
exactly 1%. Find the grades. 


PROBLEMS PERTAINING TO MENSURATION 


22. The three angles of a triangle are together equal to 180°. 
The largest angle is 4 times as large as the smallest one, and 
equal to the sum of the two smaller angles. Find the three 
angles. 

23. The perimeter of a rectangle is 330 inches. If the length 
of the longer side is doubled, the perimeter is 528. What are 
the dimensions of the rectangle? 

24. The sums of the three pairs of adjacent sides of a tri- 
angle are 180, 200, and 220. Find the sides of the triangle. 

25. The area of a trapezoid is equal to one-half the product 
of the sum of the two bases by the altitude. The altitude is 
6, and the area is 54. The lower base is 2 longer than the upper 
base. Find the upper and the lower base. 


PROBLEMS PERTAINING TO PHYSICS 


26. In Wilson and Gray’s determination of the temperature 
of the sun the Fahrenheit reading of the temperature is 5552 
more than the Centigrade reading. What is the Centigrade 
reading. See Exercise 10, Art. 48. ) 

27. It is required to find the amount of expansion of a brass 
rod for a rise in temperature of 1 degree Centigrade, also the 
length of the rod at temperature 0°. If ¢ represents the expan- 
sion, and bo the length required, it is known that b = ct + bo, 
where b is the length of the rod at temperature t. Whent = 20°, 
the length of the rod is 1000.22 ; when ¢ = 60°, the length is 
1001.65. 

28. If h represents the height in meters above sea level, and 
b represents the reading of a barometer in millimeters, it is 


Arts. 55,56] INDETERMINATE EQUATIONS 91 


known that b =k-+hm, where k and m are constants to be 
determined from observation. It is observed that at height 
120 meters, the barometer reads 751, at height 769 meters it 
reads 695. Determine k and m. Use the resulting formula 
to calculate the barometer reading for a height of 1000 meters. 

29. Two boys, Jack and Bill, wish to determine their weights 
by means of a teeter board. They find the board balances 
when Jack is 8 feet from the fulcrum and Bill is 6 feet from it. 
When Jack takes a 5-pound weight in his hand, he has to move 
4 foot nearer the fulcrum to maintain the balance. Find the 
weight of each of the boys. (See p. 22.) 

30. A weight of 400 pounds is carried by two men by means 
of a pole, at a certain point of which the weight is hung. One ~ 
man holds the pole at a distance 33 feet from the weight and the 
other at 44 feet. What amount of the weight does each man 
lift ? : 

31. Two unknown weights balance when placed 8 and 12 
inches from a fulcrum ; if their positions are reversed, 25 pounds 
must be added on the side of too small leverage in order to 
balance. What are the weights ? 


56. Indeterminate equations. In this chapter, we have been 
engaged mainly in solving systems of equations where there 
are aS many equations as unknowns. However, in Art. 49 
attention is called to the fact that a single linear equation 
in two unknowns has an unlimited number of solutions. In 
general, when the number of equations in a system is fewer 
than the number of unknowns, we have what are called inde- 
terminate equations. 

In this book, we shall consider only the case of a single equa- 
tion in two unknowns. ‘Thus, 


x+2y=4 (1) 
is satisfied by 
aio Rimes a = 2, denen 
y = 2, Y = % = 1, ) 


92 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 


It is sometimes desirable to solve such an equation for inte- 
gral values or for positive integral values of the unknowns. 


EXERCISES AND PROBLEMS 


1. Show that the equation 3x + 6y =5 has no integral 
solutions. 


SOLUTION : Since 3x + by = 5, (1) 
32 = 5 — by, (2) 
and z= 3 — 2y. (3) 


When integral values are assigned to y, we get no integral values for z. 


2. A farmer spent $386 buying two kinds of sheep, the one 
kind costing $3 per head and the other kind $5 per head. How 
many of each did he buy ? 


SoLuTion : Let, x = number of sheep at $3 and y the number at $5. 


Then 3a + Sy = 36, 
3x = —dy + 36, 
5 


To give integral values greater than 0, y can take values 3, and 6. 


Hence, 7 head at $3 and 3 head at $5, 
or 2 head at $3 and 6 head at $5. 


3. How many whole watermelons at 35 cents and whole 
cantaloupes at 15 cents can be purchased for $2.10? 

4. Separate 37 into two parts one of which is divisible by 
5 and the other by 11. 

5. How many selections of coins can be made to pay a debt 
of 60 cents with quarters and dimes ? 

6. I desire to weigh a mass of 15 ounces by means of two- 
ounce and three-ounce weights placed in the same pan. How 
can this be done? 

Historical note on indeterminate equations. It may be of interest to 


note that the early work in algebra centered on indeterminate equations 
rather than on determinate equations. The writings of the Greek alge- 


Arr. 56] HISTORICAL NOTE 93 


braist Diophantus of Alexandria (about 350 a.p.) were devoted largely to 
the solution of indeterminate equations. He did not, however, devise 
general methods for the solution of indeterminate equations. Each prob- 
lem had its own peculiar method, and it took great ingenuity to devise 
solutions for the many particular cases. 

The credit for the invention of general methods of treating indetermi- 
nate equations belongs to the Hindus. The Hindus set the problems of 
indeterminate analysis in a different form from that of the Greeks. The 
object of the former was to find all possible integral solutions while Dio- 
_ phantus was content with a single rational answer. 


CHAPTER VII 
RATIO, PROPORTION, AND VARIATION 


57. Ratio. The ratio of a number a to a number 60 is the 
a 
b 
written a: b. 

It is clear from the above definition that any ratio is a frac- 
tion and any fraction may be regarded as a ratio. Thus, 


quotient - obtained by dividingaby 6. The ratio a to b is also 


2 C ae 
2 9’ and q are ratios. 

38. Ratios involved in measurement. It is good usage 
and often convenient to speak of the ratio of two quantities if 
they have a common unit of measure. Thus, the ratio of 6 
feet to 2 feet is . 

To measure a quantity is to find its ratio to a given unit of 
measure. Thus, when we say a bar is 3 yards long, we mean 
that the ratio of the length of this bar to that of the standard 
yard is 3. 


EXERCISES 


Express the following ratios as fractions and simplify : 


1.° 6:: 8. 5. (1-4): (142); 

74) 5 haan 6. 10 feet : 6 feet. 

3. 22:8. 7. 37°53. 

4, (2? — y*): (@—y)? 8. (x? — y*): (u + y). 
1 1 


ek Pe yt Se 4 2 


Arts. 58, 59] Se ePROPORTION 95 


10. Separate the number 168 into two parts in the ratio of 
3:4, 
_ 11. Two numbers are in ratio 2 to 5, and if 1 is added to 
each number, they are in ratio 3 to 7... Find the numbers. 
12. Separate the number 168 into three parts that are to each 
other as 1:2: 4. 
Hint: Let x, 22, and 4x be the parts. 


59. Proportion. A proportion is a statement of the equal- 
ity of two ratios. Thus, 


er as he 
Gee 
is a proportion and is often written 


Ore OO: 
Itis read “ ais to b as cis to d.” 

The four numbers a, b, c, and d are said to be in proportion, 
a and d being called the extremes and 6 and ¢ the means of the 
proportion. 

EXERCISES 


Find the value of x in the following proportions : 





x 7 3 - 
i; een 10. 4, riemt 
Ete. a1 
2. 10: 5. Ca as 
Smet i2= 2:3 6. 4:7 =2: 14 
a Cc 
7. If ad = bc, show that 5 a 
Meote show that © = ¢. 
a C 
Me now that + ore 
as Ch ae OL a C 


96 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 


10. He = then xz is said to be a fourth proportional to 


a,b, andc. Find a fourth proportional to the following sets of 
numbers : 


(a) 5, 10, —6. 
(6) 4,3, —12. 
(c) 4, —6, —5. 
Liceit : = ; then x is said to be a mean proportional between 


aandd. Find the mean proportional between the following 


sets of numbers: 
(a) 4 and 9. 


(b) +3 and +48. 
(c) —3 and —48. 


PP kee °, x 1s said.to be a third proportional to a and Db. 


b 
Find a third proportional to the following pairs of numbers : 
(a) e2./83 
(b) 2, —3. 
(CMe aelals 
(d) —7, —11. 


13. Find two numbers in the ratio of 3 to 4 whose sum is 28. 


14. Find two numbers in the ratio 3 to —4 whose sum is 
=aNGY 
15. What number must be added to each term of the fraction 


1 2 
ao a fr ae 
i0 to give a fraction equal to 5 


16. What number must be added to each term of ; to: 


id pe 
=e 
obtain 3! 
Write as proportions the following : 


17. OS = 2 < 28 19..33:5 =a ae 
18. 5-8=4 > 1 20.. ab)= 4aas 


_ Arts. 59, 60] 


21. a? —b- =3.- 5. 
fee — or + 2: = ab. 
23. xy = 20. 
24. ~ = ab 
c 
25. ab = cz. 
Solve the following for x: 
BUree se 0: SS. 
31. 7:4 = 28:2. 


2 


ete 2) = 2: 


60. Properties of proportions. 


86. 4:2 = 


at 
3° 
x 


-10 — 2. 


PROPERTIES OF PROPORTIONS 


26. 4% =6- 7. 
27. a = 0b. 
Soturion: © = 2 

Tew 
28. 4:= 2% 2 
20s Tit 


33h Oe =a ar. 


34. rit eae 
x 


30S eet = L382: 


oi. () — =): (1-4-2) = (2 — 2): (2: +:2). 


97 


I. In any proportion, the product of the means equals the 


_ product of the extremes. 


For, in 


3 
b 


) 


when we multiply each member by bd, we obtain 
ad = DG: 


Hl. 1 = ©, show that 


b 


Hint: Multiply both members by °. 


d 


That is, in a proportion the means may be interchanged with- 
out destroying the equality. 

In this case, the second proportion is said to be obtained 
from the first by alternation. 


Tiss 


Hint: Divide the members of 1 = 1 by the members of the equation. 


b 


C 


d 


» show Pare = 2 
Hoes 


In this case, the second proportion is said to be obtained 
from the first by inversion. 


98 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 


is ahow that oe 


C 
LY, If; 7 5 7 








Hint: Add 1 to each member and obtain = +1= at i; 


In this case, the second proportion is said to be obtained 
from the first by composition. 
GeaeG a—b cd 

V. lf; = ae then ? = d 


Hint: Subtract 1 from each member. 








In this case, the second proportion is said to be obtained 
from the first by division. 




















pn. Ge GD: Cad 
I]. If- = -— then = . 
eg HP rire eh a 
Hint: Divide members of “ ne eb: +s in IV by the members of 
a0. 6 a 
pak aa in V. 
O.--.b° =e ae 








In this case, the proportion is said to be ob- 


poe ee a 


; a C 

tained from > = = 

bad. 

In this connection, instead of the words “composition,” “division,” 

“composition and division,’ some authors use the words “addition,” 
“subtraction,” and “addition and subtraction ”’ respectively. 


by composition and division. 


EXERCISES 
From the following proportions, find the proportions that 
arise by alternation, inversion, composition, division, and com- 
‘position and division. 


Loe) 4 3's: 4.0: > 2 
Pes Seo 5. 2:24 [See 
2 2 
3 9 
8. 7:5 = —2l eae 6. («+ y):(e—Yy) = oem 


Art. 60] EXERCISES AND PROBLEMS 99 
ia@26)=¢:d, prove 











a ¢ a C 
BoE Od’ ota pioe sy 
a@+2b c+2d 10 a+3b c+3d 
ee 25s co — 2d Sees ODAue Se OF. 
Lt en QE Ca Oe OAT Oy 
LT If; 7 UO ree rey, re lis F 
Hint: Let = 5 = 7 =k, then a = bk, ¢ = dk, ¢ = fk. 
id d3 a% 
ile If bi = be = bs = ha show that 


Atra+rd+d 


bi tbe +bs +b, by 


PROBLEMS INVOLVING SIMILAR FIGURES 


13. The sides of a triangle are 3, 4, and 5. In a similar 
triangle the shortest side is 5. What are the other sides? 
(Fig. 26.) 

Similar figures are figures of the same shape. (Figs. 
26, 27.) In two similar figures any two of the sides of 
one are proportional to the two corresponding sides of the 
second. The areas of similar figures have the same ratio 
as the squares of corresponding sides. 

14. The sides of a triangle are 9, 10, and 11. 
If the shortest side is lengthened one inch, what 
are the increases in the length of the other two 
sides in order to make the new triangle similar to the old? 

15. In Fig. 27 the sides of the larger figure are 1, 2, 3, 4, 5. 
The longest side of the smaller figure is 3. What are the lengths 
of the other sides? 

16. The area of a triangle whose base is 12 inches is 60 square 


5 
Fig. 26 


100 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 


inches. If the area of a second triangle similar to the first is 
135 square inches, what is the base of the second triangle ? 

17. A post 4 feet high, 20 feet from a 
street light, casts a shadow 7 feet long. 
What is the height of the light on the lamp 
post ? 

18. If in a map the distance between two 
points 450 miles apart is 4 inches, what is the 
distance between two cities which are 5% 
inches apart on the map? 

19. The perimeter of a triangle is 78 inches, and the longest 
side is 9 times the shortest, while the third side is a mean pro- 
portional between the other two. Find the sides of the triangle. 





Pige- 2 


PROBLEMS INVOLVING SIMPLE MACHINES 
20. The arms of a lever are 6 and 43 feet. What weight on 
the end of the short arm will just balance a weight of 100 pounds 
on the end of the long arm? 


For principle of the lever, see p. 23. The formula there given may be 

written as a proportion 
W ie fw. 

21. A pump handle is 4 feet long and works on an axis 4 
inches from the pump rod end. What force is required at the end 
of the handle to overcome a pump rod resistance of 20 pounds ? 

22. Where must the support of a seesaw board 14 feet long 
be placed, in order that two children weighing 60 and 75 pounds 
may balance? 


23. A crowbar 6 feet long is used to lft a stone weighing 


800 pounds. The fulerum of the crowbar is 6 inches from the 


end of the bar. What force must be applied at the end of the 


bar to lift the stone ? 

24. The length of the handles of a wheelbarrow is 44 feet. 
The center of a load of 200 pounds is 20 inches from the axle 
of the wheel. What force is required to lift the load ?° 


A Sila laigerion Papas ain 


Art. 60] PROBLEMS 101 


25. In a wheel and axle, the radius of the wheel is 10 inches 
and the radius of the axle is 3 inches. What force applied to 
the rim of the wheel will wind up a weight of 140 pounds? 


The wheel and axle (Fig. 28) consists of a wheel and cylinder fastened 
together and turning about the same axis. The force F’ applied at the 
circumference of the wheel to lift the weight W by a 
rope wound around the cylinder depends upon. the radii ry 
r of the cylinder and 7’ of the wheel according to the ae 
formula, (f) 


fo yo ort Fe 
_ 26. The crank to a well windlass is 20 inches F 
long and the cylinder upon which the rope is 
wound is 6 inches in diameter. What force is 
necessary to lift a bucket of water weighing 60 
pounds? 

27. After the cylinder in Problem 26 has been closely wound 
with rope and a second layer of rope has begun to be wound 
on the cylinder, how much is the force increased to lift 60 
pounds if the rope is one inch in diameter ? 

28. A brakeman pulls with a force of 160 pounds on a brake 
wheel 18 inches in diameter to wind the brake chain about the 
axle of the wheel. If the axle is 4 inches in diameter what is 
the pull on the brake chain ? 





Fig. 28 


29. A safe weighing 900 pounds is pushed up an inclined plane 
by means of rollers. What force is necessary to keep it from 
F slipping back if the inclined 
plane is 10 feet long and raised 

2 feet at one end. 


The force F necessary to hold a 
weight W on an inclined plane (Fig. 
Fie. 29 — 29) is given by the proportion 


F:W =h:l, ; 


where | is the length of the plane and h the height through which one end 
is raised. No account is here taken of friction. 


102 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 


30. Find the force necessary to prevent a barrel weighing 
350 pounds from rolling down a plane 21 feet long, one end of 
which is raised 4 feet. | 

31. If a boy can exert a maximum force of 100 pounds, what 
is the maximum height of the end of the inclined plane in 
Problem 30, so that the boy can handle the barrel ? 


61. Variation. In Chapter V we have seen that if y is a 
function of x, written 
ya F(x), 
then in general y changes when x changes. We may say that 
y varies when «x varies, but the word ‘“‘varies” has come to 
have a more restricted meaning when used in this connection. 
Each of the statements 


“y varies as 2,” 
‘“‘y varies directly as x,” 
‘“y is proportional to 2,” 
‘“‘y is directly proportional to 2,” 
means that y equals the product of « by a constant. That is, 
U= kx. 
The constant k.is called the constant of variation. 
The expression “y varies as «”’ is sometimes written 
Ya L. 
The area of a circle varies as the square of its radius. That is, 
) A = kr?, 


if A represents the area and r the radius. With our restricted 
‘meaning of the word “ varies,” it is not correct to say that 
the area of a circle varies as the radius, for, in the equality 
A =k-r, k isnot a constant for different values of r. 

If a train moves with a uniform speed, the distance s travy- 
ersed varies as the time ¢t. That is, 


San 


Arts. 62, 63, 64] VARIATION 103 


62. Inverse variation. Each of the statements 


‘“‘y varies inversely as 7,” 
iz3 = by me ay 2 ) 
. “y is inversely proportional to 2, 


means that y is equal to the product of the reciprocal of x and 
a constant. That is, 


Y= 


Thus, the volume of air in the cylinder of a bicycle pump varies inversely 
as the pressure on the piston. That is, 


te 
p 
if V represents volume and p pressure. 


63. Joint variation. The statement ‘“z varies jointly as x 
and y”’ means that z equals the product of x, y and a con- 
stant. That is, 

gered an Tys 
The distance which a train, moving with a uniform speed, 
travels varies jointly as the speed and the time, or 


d= kut, 


where d is the distance covered, v the speed, and ¢ the time. In 
this case k = 1, if v and d are measured with the same unit of 
length. 


64. Combined variation. The statement ‘“z varies directly 
‘as x and inversely as y”’ means that z varies jointly as 2 and 
the reciprocal of y. That is, 
ka 
g=—: 

Y 
If T varies directly as x, directly as the square of y, inversely as 
w and inversely as the cube of v, we have 


T =p. 
w v8 


104 RATIO, PROPORTION, AND VARIATION [Cuap. VII. — 


The attraction F of any two masses m and mg for each other 
varies as the product of the masses and inversely as the square 
of the distance r between the two bodies. That is, | 


kmume 
— , 9 e 


PF 





EXERCISES AND PROBLEMS 


Reduce each of the following statements, Exercises 1 to 7, to 
an equation: 
1. y varies as x, and y = 34 when z = 2. 
SOLUTION : y = ke. 
Since y = 34 when z = 2, we have 
34 = 2k, ork = 17. 
Hence, hy 2 
2. T is directly proportional to ¢, and ¢ = 3 when T = 27. 
3. n varies inversely as m, and n = 39 when m = 3. 
4, S varies as ??, and S = 144 when ¢t = 3. 
5. A varies jointly as B and C. When B = 3 and C =4, 
it is found that A = 14. 
6. P varies directly as q and inversely as r?. When gq =1 
and r = 3, it is found that P = 27. 
7. The surface S of a sphere varies directly as the square of 
the radius r. The surface of a sphere of radius 1 is 47. 
8. If x varies as y, and if y = 1 when zx = 8, find x when 
ge): 


SOLUTION : x =ky. 
Since y = 1 when z = 3, we have 
eH open Lg aes 
or t= oy. 


When y = 5, we find x = 15. 


9. Ifa varies as y and if « = 3, when y = 6, find y when 2 = 7. 
10. If a varies inversely as b, and if a = 3 when 6b = 1, find 
a when 6 = 8. 
11. If 2 varies jointly as z and y, and if 2 = 120 when a = 2 
and y = 3, find z when x = 3 and y = 3. 


Art. 64] PROBLEMS 105 


12. If z varies directly as x and inversely as y, and if z = 120 
when x = 2 and y = 3, find z when z = § and y = 3. 

13. If x varies directly as the square of y and if x = 4 when 
y = 1, find x when y = 4. 

14. The safe load of a horizontal beam 10 feet long supported 
at both ends varies jointly as the breadth and square of the 
depth. If a 2 by 6 white pine joist 10 feet long safely holds up 
800 pounds, what is the safe load of a 2 by 8 joist of same length? 

15. Write in the form of an equation the law: The safe load 
w of a horizontal beam supported at both ends varies jointly 
as the breadth b and the square of the depth d and inversely 
as the length / between supports. 

16. A beam 15 feet long, 3 inches wide, and 6 inches deep ~ 
when supported at both ends can bear safely a maximum 
load of 1800 pounds. What is the maximum load for a beam 
of the same material 10 feet long, 2 inches wide and 4 inches 
deep ? 

17. What is the safe load for the second beam mentioned in 
Problem 16 if it is turned so that the width is 4 inches and 
depth 2 inches ? 

18. Write in the form of an equation the law: The crushing 
load of a solid square oak pillar varies directly as the fourth 
power of its thickness ¢ and inversely as the square of its 
length l. 

19. If a four-inch oak pillar 8 feet high is crushed by a 
weight of 100 tons, what weight will crush a pillar half as high 
and 6 inches thick? (See Problem 18.) 

20. What weight will crush a four-inch oak pillar 4 feet 
high ? : 

21. The deflection d of a rectangular beam of a fixed length 
varies inversely as the product of the breadth 6 and the cube of 
the depth d. Write this statement in the form of an equation. 

22. In the formula 

kbd 





s = ) 


106 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 


s denotes the strength of a rectangular beam, b, d, and I, the 
breadth, depth and length respectively, of the beam, and k is a 
constant. Translate the formula into English using the terms 
of variation. 


PROBLEMS INVOLVING MOTION 


23. The number of feet a body falls varies directly as the 
square of the number of seconds occupied in falling. If the 
body falls 16.1 feet the first second, how many feet will it fall 
in 6 seconds? 

24. How far will a body fall during the sixth second ? | 
25. The velocity of a falling body at any time varies directly 
as the number of seconds occupied in falling. What is the 
velocity at the end of 6 seconds if the velocity at the end of the 

first second is 32.2 feet per second ? 

26. An object dropped from a balloon strikes the ground in 
7 seconds. At what velocity does the object strike the ground 
and what is the height of the balloon when the object is 
dropped ? 

27. A wrench is dropped from an automobile at a height of 
3 feet while the automobile is traveling at the rate of 70 miles 
an hour. How far does the automobile move while the wrench 
is falling ? 

28. The time for one vibration of a pendulum at a given 
place varies as the square root of the length of the pendulum. 
In Chicago a pendulum 4 feet long requires 1.1 seconds for a 
vibration. What is the time of vibration of a pendulum 1 foot 
long ? 

29. What is the length of a pendulum which vibrates every 
second at Chicago ? 

30. A weight is suspended by a wire 94 feet long. What is 
the time of one vibration at Chicago ? 

31. A pendulum supposed to vibrate every second registers 
90,000 vibrations in 24 hours. How much must the pendulum 
be lengthened ? 


Art. 64] PROBLEMS 107 


PROBLEMS INVOLVING PRESSURE 


32. The volume of a gas enclosed in a vessel varies inversely 
as the pressure upon it. Twenty-four cubic inches of air under 
a pressure of 100 pounds will have what volume when the pres- 
sure is decreased to 50 pounds? 

33. If a toy balloon contains 150 cubic inches of gas when 
under a pressure of 15 pounds per square inch, to what size 
will it shrink if subjected to a pressure of 45 pounds per 
square inch? (See Problem 32.) 

34. The pressure of wind on a sail varies jointly as the area 
of the sail and the square of the wind’s velocity. When the 
velocity is 15 miles per hour, the pressure on a square foot is 
1 pound. What is the velocity of the wind when the pres- © 
sure is 10 pounds per square foot ? 

35. The pressure of gas in a tank varies jointly as its density 
and its absolute temperature. When the density is 1 and the 
temperature 300°, if the pressure is 15 pounds per square inch, 
what is the pressure when the density is 2 and the temperature 
290° ? 


CHAPTER VIII 
EXTENSION OF THE NUMBER CONCEPT 


65. Integers and fractions. The first numbers studied in 
school are the numbers called positive integers. With these 
numbers the pupil is able to add and to multiply two numbers 
together and to subtract a smaller from a larger. He under- 
stands the answer, for it is always a positive integer. But 
division cannot always be performed if we have only positive 
integers; for example, the division of 4 by 3 is set aside as 
impossible. 

A student knowing only the positive integers might learn to 
solve equations of some. kinds. For example, the equations 
4—2z =0, 32 — 6 = 0, would give no difficulty, but the solu- 
tion of the equation 2x — 5 =0 would be declared impossible, 
for the student knows no numbers which will satisfy the equa- 
tion. He might call the result imaginary. 

With the set of positive integers our elementary mathematics 
goes little further than counting; even the simple operation of 
measuring cannot be performed. For this reason early in his 
school life the number system of the pupil is increased by add- 
ing to it the numbers which we call, fractions. These new 
numbers, while having no meaning in counting, have very 
definite practical applications. | 


EXERCISES 


Assuming that we know only positive integers and zero, show 
which of the following exercises are possible of solution. 

Solve: 

1; ¢-— 3: = 0. 8. 37 + Dae 

2. 32 — 15 =e 4. 2xr+/7=0, 


_ Arts. 65, 66] NEGATIVE NUMBERS 109 


6. 52 + 42 = 0. . 8. 27 — 1 = 0: 
beer — 0 = (0. Geer 4 re i). 
fe oe + 2 = 0. 10. 2? — 32 = 0. 


Discuss the results of the following problems : 


11. Three boys out hunting shot altogether 21 rabbits. Two 
of the boys shot an equal number, but the third boy shot 2 more 
than either. How many rabbits did each shoot ? 

12. Three boys gathering nuts get altogether 21 bushels. 
Two of the boys gather the same amount, but the third boy 
gathers 2 bushels more than either of the others. How many 
bushels did each gather ? 


66. Negative numbers. The numbers consisting of posi- 
tive integers and fractions are sometimes called the numbers 
of arithmetic. With this set of numbers, the operations of 
addition, subtraction, multiplication, and division, can be per- 
formed with but one exception. This exception is the sub- 
traction of a larger number from a smaller, which is declared 
impossible. For example, there is no number in the set 
which answers the question, ‘What number added to 4 
gives 3.” If astudent, knowing only this set of numbers, knows 
how to solve an equation, he would declare the solution of the 
equation 272 + 3 =0 to be impossible, or he might say the 
answer is imaginary. 

In reading a thermometer, in debits and credits, in latitude 
and longitude, a physical meaning is found for the subtraction 
of a larger number from a smaller. The introduction of new 
numbers called negative numbers is one of the purposes of the 
study of algebra. With positive and negative integers and 
fractions the student is enabled to solve any problem which 
reduces to an equation of the first degree. The equation 


ax +6b=0 


of Art. 22 has always a solution among these numbers. 


110 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. 


EXERCISES 


Assuming that we know only positive numbers, show which 
of the following exercises are possible of solution. 


1. 22 —1 =. 6. x? — 54 +6 = 0. 
2. BLS DF, 7 v—-x-2=0. 
$3. 2.42 = 4. 8. 2? — 32 4+2=0. 
4. 4¢4+9=1. 9. 2277 + 37 — 2 =O: 
6.52 = 2a: 10. 2? 4+ 1-="0: 


How many answers are there to each of the following prob- 
lems? 

11. Two thermometers differ by 3 degrees in reading. If 
. the reading of one thermometer is doubled it equals the reading 
of the other. What are the readings of the two thermometers ? 

12. The ages of two children differ by 3 years. The age of - 
one child is double the age of the other. What are the ages 
of the children ? 


67%. Irrational numbers. The quotient of two integers is 
called a rational number. Since an integer may be written 
as a fraction with 1 as a denominator, the numbers discussed 
in Arts. 65 and 66 are rational numbers. 


For example, 3, —6, 4+, .67, —} are rational numbers. 


In operations with radicals we often deal with irrational 
numbers. Irrational numbers cannot be expressed as the 
quotient of two integers. 


For example, V2, Vi, V4 are irrational. 


A student having only rational numbers at his command 
would find it impossible to obtain the length of a diagonal of a 
unit square. In other words, he would not be able to solve 
the equation zx? = 2. A further extension of the number sys- 
tem to include irrationals is then necessary. 


_ Arts. 67, 68,69] REAL AND IMAGINARY NUMBERS II11 


EXERCISES 

Represent by points on a straight line the following numbers : 

1. 4, —3, —3, §, 0. 
2. V2. 


SoLuTion: Construct a square of side 
1 (Fig. 30). The length of its diagonal is 
V2. Laying off the length of this diago- 
nal from Owe find a point on the line 





representing ~/2. 


3. V8. 

4.2+V2, 2-v2, -2- v2, -24+ v2. 

Bree + V/8, —3 + V8. 

Arrange the following sets of numbers in ascending order. 
6. V2, -4,2 — V2, -V8, 3, 2vV2. 

Pert 4) —/2, —2, 4, —3V2, 


68. Real numbers. The four classes of numbers which we 
have discussed, that is, positive integers, fractions, irrationals 
and negative numbers, together with zero are classed together 
under the name real numbers. These numbers have the im- 
portant property that they may all be represented by points on 
the same straight line. 


69. Imaginaries. We have seen that man found it desirable 
to invent fractions, irrationals and negative numbers. In just 
the same way it is found convenient to invent still another kind 
of number. The square of any real number, positive or nega- 
tive, is a positive number. The square root of a negative 
number cannot then.be a real number and is given the name 
imaginary number. It is not represented on the same line 
with the real numbers. 

The term “ imaginary numbers ”’ is here used in a technical 
sense. The numbers are imaginary in the same sense that a 


y] 


112 EXTENSION OF NUMBER.CONCEPT ([Cuap. VIII. 


fraction, a negative number or an irrational number is imagi- 
nary for a person knowing only the positive integers. 

It is desirable to include these imaginary numbers in our 
number system; for otherwise we could not solve such an 


equation as 
24+1-=0. 


The number 1/—1 bears somewhat the same relation to the 
system of imaginary numbers that the number 1 does to the real _ 
numbers. Any real number may be written with 1 as a coeffi- — 
cient; forexample, 4 = 1-4. Any imaginary number involves 
the product of a real number and ~/ —1. 

Thus, we may write 








Af HA DA: 
/ 5 SAD ee 
EVE va 


Daf SO) ee) RAY 


The number V —1 occurs so often that a special symbol is 
used for it; that is, 


EXERCISES 


Raa, 
i. 


Write the following imaginary numbers in terms of 7 
1. /—5. 
Sotution: V-5 = V5- -1=V75-vV -1 =iv5. 








ave 6. V—ae 

S20 A 0! 7. Voie 
: ae ar 

A/S 8. \/ re. 

5. =. 9. V-07- oe 


10. W-1 — 42. 


Arts. 70, 71] IMAGINARY NUMBERS 113 


70. Powers of i. Since any imaginary may be written in 
terms of 7, it is important to consider the powers of 7.. We 
operate with 7 as with any other letter and remember that 2? 
is to be replaced by —1. Hence, for powers of 7, we have 


er |) 
aa a 
pee (ay) "1: 


SAS 0 hl ee 
v= ts 742 = —), 
and so on, repeating the numbers —1, —7, +1, +7. 
71. Products of imaginaries. If imaginary numbers are 


written in terms of 2, the product of any number of imaginaries 
is easily found. 


Example. Find the product of —3, /—5, and /—7. 
SoLUTION: V/-3 =iv3, V—-5 =iv5, V-7 =iv’7. 
V8 V5 VAT =iv8 1 V5 iT 
= OVS 5 «V7 
= —iV/105. 

In general if ~/—a and —6 are two imaginaries we have 
Va: V—b= iva: ivb =?Va- Vb = -Va- Vb = —Vab, 
ye ae /-a-/—b = — Vab. 

| EXERCISES 

1. Continue the table of Art. 70 for 27, 28, 79, 7, a. 


Find the following products : 








Pewee... 4/ —3. 1. VSL een 
Bape a. 8. (/—2)3. 

ee 4/ —10: . 9. (V5) 

5. s/he n/— 16. 10; (\/ 4) 

Bn cay nt 4/ — 11; Et (/ —3) Vas 


12. (V—-4)'V—§. 


114 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. 


72. Complex numbers. The sum of a real number and an 
imaginary number is called a complex number. Thus, 1 +7, 
2 —\/— 3, 5 + 37 are complex numbers. In general, a complex 
number has the form 

a + 2b, 
where a and b are real numbers. The expression “ imaginary 
numbers ”’ is often used to mean complex numbers, and the 
term pure imaginary is applied to numbers of the form 7b. 

We shall find such numbers in solving quadratic equations. 


‘ 


For example, 2 + ~/—38 satisfies the equation 
xv? —4244+7 =0, 

as shown by actual substitution. 

Thus, 
(24/7 —3)? -42 + J7-3) +7 =444/-3 —38 = See 

73. Operations with complex numbers. The ordinary op- 
erations of algebra may be performed with complex numbers 
by operating with 7 as with any other letter and remembering 
that 7? = —1. 

Example 1. Add 6 + —3 and 3 -— W-2. 

In the 7 notation we are to find 

(6 +7173) + (8 —iv2) =6 +1V3 43 -iv?2 =9 +7(vV3 —- V2). 


Example 2. Multiply 6 + /—3 by 3 — V—2. In the 7 notation 
(6 +iV3) (8 — iV2) = 18 - 6iV2 + 3iV3 — 23-2 
= 18 —1(6/2 — 3V3) — (-1)V6 = 18 + V6 —71(6V2 ae 
74. Conjugate complex numbers. Complex numbers which 
differ only in the sign of the imaginary parts are called con- 
jugate numbers. Thus, 3 + 22 and 3 — 21, a+ 7b and a — 1b 
are conjugate. Since 
(a + 2b) + (a — 2b) = 2a, 
(a + 1b) — (a — 2b) = 20d, 
(a + 2b) (a — 1b) = a? + B?, 
we see that the sum and product of two conjugate complex — 
numbers are real numbers, but the difference is an imaginary 
number. 


Art. 76] REVIEW EXERCISES OF PROBLEMS 119 


REVIEW EXERCISES AND PROBLEMS ON CHAPTERS V-VIII 


1. If x varies inversely as y, and when x = 6, y = 2, what is the value 
of x for y = 3? 
2. If x varies as y, show that x + y varies as y. 
3. Determine x and y from 
0.52— sy =0.15, 


4¢ +0.2y = 2.41. (STANFORD) * 
4. Solve a i = 4, 
ay 
i ra = 6. | (YALE) 
ey), 


5. Find a mean proportional between (@ + y)? and (a — y)?. 








6. Find a third proportional to - + : and - 
7. Evaluate li 2ar3 
4 5 6 
(Pan 
8. Solve the equation La 2B EDs 
pe ee 
| See: 








9. The weight of an object above the surface of the earth varies in- 
versely as the square of its distance from the center of the earth. An ob- 
ject weighs one pound at the surface of the earth. What would it weigh 
1500 miles above the surface? (Take 4000 miles for the radius of the earth.) 

10. The weight of an object below the surface of the earth varies directly 
as its distance from the center of the earth. An object weighs one pound 
at the surface of the earth. What would it weigh at a point 500 miles 
below the surface ? 

11. Simplify (a — ib)? — (a + 2b)?. 

12. Represent graphically on the same diagram the numbers 2 — 31, 
ya | 


Be Aseely tae S, sl gma V/2— in/2. 
13. Solve . 3ax — 2by =, 
a®*z + b’y = 5hbe. (SHEFFIELD) 


14. If the base of a triangle is constant, show that the area varies 
directly as the altitude. 





* Institution that gave the question in an entrance examination. 


120 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. 


15. Show that the radius of a circular cylinder varies inversely as the 
square root of the altitude if the volume is constant. 

16. The price of sugar for the 21 days immediately following the open- 
ing of the European war was as follows: 4.40, 4.40, 4.40, 4.40, 5.00, 5.00, 
5.00, 6.00, 6.50, 6.75, 7.50, 7.50, 7.50, 7.50, 7.50, 7.50, 7.25, 7.25, 7.00, 
7.00, 7.00. Show the change in price by a graph. 

17. On the same sheet of codrdinate paper, represent graphically the 


: 2 
three equations y=27?+24+1,y=1l-2,y= ach 1. What can be told 


from the points where the graphs intersect each other? 
18. Give an example of two linear equations in two unknowns whose - 
graphs are parallel. 
19. Give an example of two linear equations whose graphs coincide. 
20. If a:b =c:d, prove that ma +b: ma —b = mc +d:me —d. 
21. If 7x — 42:82 — 32 = 4y — 7z: 38y — 8z, prove that z is a mean 


proportional between x and y. 
22. The following scores were made by city school children and by rural 
school children in grades from the third to the eighth inclusive, in a speed 


test in the fundamental operations in arithmetic. 


Grade 3rd 4th 5th 6th 7th 8th 











iby 5.4 | 6.6 | 9.0 | 10.3 | Tb 13.1 
Rural | 2:5 6.5 6.9 8.0 10.0 11.1 
Make graphs of these scores to show the comparative speeds. 
Lay it 
23. Solve: — pat, 
Vay 
ee 
pide ae 
i ea 
Ps + mt v5 (Mass. INstiTuTE) 


24. If one of two numbers be multiplied by m and the other by n, the 
sum of the products is P, but if the first be multiplied by m’ and the second : 
by n’, the sum of the products is P’. Find the numbers. (DarrmMoutH) 

25. Represent graphically, y = x? — 2z. (WISCONSIN) 

26. Compare the cubes of the following numbers, 

Seis) 1 1/3 
oe 2 

27. By actual substitution show that 3 + 27 and 3 — 27 de the equa- 

tion x? — 62? + 13x = 0. 


1, 


Art. 76] REVIEW EXERCISES AND PROBLEMS 121 


28. Given that z varies as the sum of two quantities one of which 
varies directly as x and the other as x?. If when x = 1, 2, 3, z = 5, 16, 
33 respectively, what is z in terms of x? 

29. Simplify (@® + 21° +21 + 72)’, if 7 = —1. 

30. Find the value of y by determinants from the equations : 

3a — 4y + 22 =1, 
2x + 3y — 32 = -1 
oa — Sy + 42 = 7. 
(COLLEGE ENTRANCE EXAMINATION BOARD) 


d 


mae — meh —,, V—-2 + 37-1. 
31. Simplify 8V —2 + V —4)(2V-3 -— V-2) ; es ea 
(ILLINOIS) 


32. The force P necessary to lift a weight W by means of a certain 
machine is given by the formula P = a + bW, where a and Bb are constants 
depending on the amount of friction in the machine. If a force of 7 pounds 
will raise a weight of 20 pounds, and a force of 13 pounds will raise a weight 
of 50 pounds, find the force necessary to raise a weight of 40 pounds. (First 














determine the constants a and 6.) (HARVARD) 
33. Solve the simultaneous equations 
ee Loh we Peele y. isa), 
—-1 1 -1 i ae 
eS Eat =i 2 20 
ea -* =-, prove that ax +by +cz is a mean _ proportional 


between x? + y? + 2? and a? + b? + c?. 
35. Find a number such that by adding it to each of the numbers, 
a, b, c, d, we obtain four numbers in proportion. (Mass. INstTiTuTE) 
36. A glass of milk and cream is sold for the same price as a glass of 
milk of twice the size. If milk costs 7 cents a quart and cream 32 cents a 
quart, what is the percentage of cream in the smaller glass? (HarvARD) 


37. Solve: “4x — 8y. = —3; 

? llx + 5y = -15. 

Draw a graph and verify your answer. (Missourt) 

38. If a:b =c:d, prove that ab + cd is a mean proportional between 
a? +c? and b?+d. — (Mass. INSTITUTE) 


39. A man invested $2720 in railroad stock, a part at 95 yielding 2 per 
cent, and the balance at 82 yielding 3 per cent; his income from both in- 
vestments is $70. Find the amount invested in each kind of stock. 

(ILLINOIS) 


122 EXTENSION OF NUMBER CONCEPT ([Cuap. VIII. 


40. Suppose the earth to be a smooth sphere 25,000 miles in circumfer- 
ence and that an iron band 1s fitted closely around the earth at the equator. 
Suppose this band to be expanded by the heat of the sun until its inner 
circumference is 25,000 miles and 1 foot in length, and that the distance 
between the surface of the earth and the band is the same at all points. 
Is it possible to insert a knife blade ;4 inch thick between the surface of 
the earth and the band? Estimate your answer and then check it. 

41. A concrete pavement is 28 feet wide. A cross section gives an arc 
of a circle with an 8-inch rise at the center. (Fig. 34.) It is necessary to 
determine the radius of this circle in order —___-———+,———__-— 
to construct the form (templet) by which dork 
the concrete is laid. Find the radius. Fic. 3¢ 

42. Solve the above problem if the width of the pavement is a and the 
rise at the center is D. 

43. The lengths of the sides of a triangle are consecutive integers. If - 
-a represents the side of middle length and A the area of the triangle, then 
A =f(a). Find f(a). 

44. If 8 yards of silk and 12 yards of woolen cost $27, and 12 yards of 
silk and 8 yards of woolen cost $28, find the price per yard of the silk and 
of the woolen. (CALIFORNIA) 

45. The radius r of one base of a segment of a sphere is one-half the 
radius of the other base. The altitude of the segment is equal to the radius 
of the smaller base. The volume of the segment and the surface of the 
zone are then functions of r._ Find these functions. (See Formulas 18 and 
19, Art. 45.) 


CHAPTER IX 
QUADRATIC EQUATIONS 


77. Typical form. A quadratic equation in one unknown zis 
an equation that can, by transposing and collecting terms, be 
written in the typical form 


ax? + br +c =0, 


where a, 6, c do not involve x, and have any values with the 
one exception that a is not zero. Since the result of multiply- 
ing the members of an equation in this typical form by any 
given number is an equation in the typical form, the a, b, c can 
be selected in an indefinitely large number of ways. 

A quadratic equation is said to be of the second degree. 

The function az? + bx + ¢ (a # 0) is called the typical quad- 
ratic function. 


EXERCISES 
Arrange the following equations in the typical form and 
select a, b, and c from the resulting equations : 


2 
1. Br +44 5-5 $241. 


SoLuTIoNn : By transposing and collecting terms, 


8 Z 
ea AD hee en pa 
at 5 +3 0, 
, 8 1 
so that Segiei 5) & = 9. 


2.93 +27 — (27? = 24 + 9 — 9x7. 
Sere Wi)? = x + 2. 
1 


es Rae 4 


124 QUADRATIC EQUATIONS [CHar. IX. 


] 
5. 4y* + oy +6 = 4 — 8y + dy’. 


6. 22432-2744 oe 


ier 22 x 
~ sats 4_—_ = —* 
7 a eR + 2 i I+ i 
2 
8. 32? —xa +k =a" + 2. 
SouuTIon : By transposing and collecting terms, 
7 
at? — x + hein = (), 


so that | a= 


9. 4? + (22-4 n)? = 1) 11. (mx +n)? + 2? = n?. 


(@ ++-1)? (eae 
12. Tg 2 
138. (k — x)? — (n+ 2)? = (m — @)?. 


10. 274+ 2nr+t=k4+ te. 


14. d*y? + y? ae (y+d)? =0. 
15. (¢+n)? — ( — nj = ni’. 


78. Solution by factoring. In solving a quadratic equation, 
it is generally best first to reduce the equation to the typical 
form. After such a reduction, if the left-hand member of the 
equation can be factored readily, the solution is very simple. 


Example. Solve (x —.3)?.= 6.— 2a. . 
Arranged in the typical form this equation becomes 
xv?—4r +3 =0. 
The factors of the left-hand member are easily found. They are x —3 
~ and x — 1, and we may write the equation in the form 


(¢ —3)(4 — 1) = 0. 
Any value of x which makes either factor zero will satisfy the equation. 
If x = 3, we have (3 —3)(8 —1) =0-2 =0. 
Again, if x = 1, we have (1 — 3)(1 — 1) = -2-0=0. 
Hence, 3 and 1 are solutions of the given quadratic equation. 


_ Arts. 78, 79] SOLUTION BY FORMULA 125 


EXERCISES 
Solve the following by factoring: 
ih a +62 + 5 = 0. 9. 377+ 4r+4+1 =0. 
Bee 3)? = 1. 10. 32? — 17x + 10 = 0. 
Sat = 025. 11. 6y? — Sy + 2 = 1. 
ace = 12. 122 287 SS 9. 
5B. (x +1)\(a@-—1)-8=0. 13. 4+ 2(42 —17) =0. 
6. x2? — nx = mn — mx. 14. 2x? — ax — a’? = 0. 
To 2? — 2nz+ n? = 0. 15. (@ +1)? = 2(2? — 67 — 3). 
8. 277+ 54 — 3 = 0. 16: a? ='Sa? — 2a: 


79. Solution by formula. Any quadratic equation may be 
solved by the process of “‘ completing the square.” 


For example, to solve 2%? — 4% — 7 = 0, write the equation in the form 


x? — 2x = ni Adding 1 to both members to make the left-hand member 


2 

7 9 
a perfect square, 22? —22 +1 = mee 1 = 5 
or (x —1)? = . 

9 3 

Extract the square root, 2-—1=+ as ate BV 2; 

3 = etae 
and | x =1+5V2, orl -5v2. 


Both of these values of x satisfy the original equation, as we see on substi- 
tuting them forz. Thus, 


x a 
; = ee _ 9 3 
2\1 rea —4\ 1 45/2 ) —7 =2\ 14+3V72+,-2) -4 1422 —7. 
2 2 4 2 
=2+6/7249 -4 -6vV2 -7 =0. 
In the same way, substituting 1 — SV 2 for x, we find 


aie ae ee 
2(1 - 5v2)? - 4(1 -5v2) -7 =0. 


126 QUADRATIC EQUATIONS [Cuap. IX. 
If we apply this method to the typical quadratic equation 
ax? + bx +c = 0, 


we obtain a formula by means of which we can write down 
the roots of an equation without factoring or conn Cue 
the square. 

Transposing c and dividing by a, the general equation be- 
comes 

5 EO c 

ee +-% = —-: 

ca a 


Add the square of one half the coefficient of x to each member 


to make the left-hand member a perfect square, i 
oi, (2)}. 2: (2 
- 90s aha 2a) aaa 
+5) uF O5 ee tO 
* pee he eee Te. 


Extracting the square root, 
bees Mb = 4ac 


eGo 2a 
—btivb? — 4ac 
or Qs) a 
2a 


If we let x, and 22 represent the two solutions of the quadratic — 


equation 
az +ba+e= 0, 





‘ —b +2/b?— 4ac 

_ we may write Ly = 2 =) 
a 
2 

cae ee; —+/b?— 4ae 4ac 
2a 


These expressions may be used as formulas for the solution 
of any quadratic equation. 


— Arr. 79] EXERCISES AND PROBLEMS 7 


Thus, to solve the equation 
2x? — 4x.— 7 =0, 





we substitute in the formulas, a = 2,6 = —4, c = —7, and find 
a Ee + Vv 16 SEE Cah) af SEA oe 1 +33, 
ee 2 7) es 3g. 
EXERCISES 


Solve the following equations by use of the formulas, and 
verify by substitution: 


1. v7 -— 4% +3 = 0. 9. 677 +2 - 12 =0. 

Zor — (x + 10 = 0. 10. 9x? = 13 — 42. 

3. 224+ 7r +12 = 0. 11. 97? + 147 +3 =0. 

4.327 + 10x = 82. 12. 3a? — 187 + 5 = 0. 

5. 2r? — 3a — 14 = 0. 13. 1077 + 92 + 8 = 0. 

6. 477+ 7744+ 3 = 0. 14. x7 + 6.51 — 5.27 = 0. 

toon — 2 —6 = 0. 15. (1 —2z)(x —2) +3 =0. 
1 3 

8. 67? + 54 +1 = 0. 16. xi + 5t —7— = 0. 


17. y? + 0.9y = 1.2. 


- Solve by any method : 
18. 16(1 — x)(1 +2) — 92? = 0. 
19. 22° +974+14=0. © 22. x(x — 2) =5 4+ 32. 
20. 22 +7x —-9 =0. 23. (2x — 5)? — (a — 6)? = 80. 
21. 32? + 1.03¢ + .1 = 0. 24. 3x7 = 332 — 90. 
25. (x —6)(x —5) + (& - 7)(a — 4) = 10. 
26. Divide the number 29 into two parts such that. their 


product is 198. 
27. A square is cut from a rectangular piece of paper. The 
remainder of the rectangle has the shape and dimensions shown 


128 QUADRATIC EQUATIONS [Cmap. IX, 


in Fig. 35. What is the side of the square if the area of the origi- 
nal rectangle was 100 square inches. 
28. The product of two consecutive numbers is 182. Find 


the numbers. 
29. Three times the product of two con- 


secutive even integers exceeds ten times 
their sum by 60. Find the numbers. 

30. The sum of two adjacent sides of a 
rectangle is 28 inches, and the area of the 
rectangle is 195 square inches. What are 
the dimensions of the rectangle ? 

31. One side of a rectangle is 3.5 inches longer than the 
other. The area of the rectangle is 181 square inches. What 
are its dimensions ? 

32. A picture whose dimensions are 56 x 60 inches has a frame 
of uniform width whose area equals that of the picture. Find 
the width of the frame. 

33. Within a rectangle whose sides are 30 and 40 inches, a 
second rectangle is drawn so that its sides are everywhere 
equally distant from the sides of the given rectangle. The 
area of the second rectangle is one-half that of the first. Find 
the perimeter of the second rectangle. , 


y PHIG?H 85 


8@. Special or incomplete quadratics. If b orc is zero in 
the quadratic equation 
ax? + bx +c = 0, 

the equation is said to be incomplete. 


I. When c = 0, ax? + bx = 0 is the typical form of the equa- 
tion. We can always write this equation in the form : 


z(ax + 6) = 0. 


=p : 
Hence, the roots are 0 and ee Conversely, if 0 is a root of a 


quadratic equation 
ax? + bu +c = 0, 


Arr. 80] SPECIAL QUADRACTICS 129 


then a-0+6-0+c=0. 
That is, c= 0. 

Therefore, a quadratic equation has a root equal to zero when 
and only when the equation has no known term. 

II. When 6 = 0, ax? + c = 0 is the typical form. 
In this case, Case ya 
Conversely, if the roots of a quadratic equation are arithmet- 


ically equal, but opposite in sign, there is no term containing 
x in the first degree ; for if +7 and —r are both roots of 


ax? + br +c = 0, Cl) 
we have ar? + br +c = 0, G2) 
and ar? — br +c = 0. (3) 
Subtracting (3) from (2), 2Qbr = 0. 


Since it is given that r is not zero, it follows that b = 0. 


Hence, a quadratic equation has two roots arithmetically equal 
but opposite.in sign, when and only when the term in x vanishes. 

III. When b = 0, c = 0, the typical form is ax? = 0. Both 
roots of a quadratic equation are equal to zero when and only when 
the known term and the term in « vanish. 


Solve : EXERCISES 
1. (a — 6)? = 72 — 122. Santi er*s d. 
x? rs (x + 3)( — 3), 2 ne 
2. 9 = 3 4, x? — 32 = 0. 


5B. (@ +1)? + (& + 2)(@ + 8) = 7. 
6. (5 — x)(a — 6) + 30 = 0. 
Determine k so that each of the following equations shall have 
one root equal to zero: 


1. of oe + 3k —.15 = 0. 8. 2? +72 1 ee: 
9. 2y? — 5 +k — 4k? = 0. 


130 QUADRATIC EQUATIONS [Cuap. IX, 

Determine k and m so that each of the following equations 
shall have two roots equal to zero: 

10. 5x2 + 3ka — Q2mz — 5a +k+m—5=0. 
11. 247? 4+ 4k?a + 8ka — 38m —1 = 0. 

Determine k so that the roots of the following equations may 

be arithmetically equal but opposite in sign: 
12. 277 ++ 4+ 2kx +1 =0. 
13. k*x.— 9x — 3x? = 14. 
14. 2? +4 7k*z — 2kz — 8 = 0. 

81. Formation of equations with given roots. It is some- 
times necessary to form a quadratic equation with given roots. 
The following theorem and its converse enable us to do this 
very easily. 

TuHEorEM. If r is a root of the equation 

ax? + bx +c = 0, 
then «a — r is a factor of the left-hand member. 


If r is a root of the equation, we have 
ar? +br+c=0. 
Then aav*?+bxa+c = az? + bx+ec-— (ar? +br+c) 
= a(z? — r?) + b(@ — r) 
(x —r)(ax + ar +b). 


Hence, x — r is a factor of ax? + bx + €. 


ConVERSE THEOREM. If #—vr is a factor of the left-hand 
member, then r is a root of the equation. 

If x — r is a factor of ax? + bx + c, then the substitution of r 
for x makes the factor x — r vanish, and 7 is a root of 


az? + br4+c=0. 


Arts. 81, 82] NATURE OF ROOTS 131 


EXERCISES 


Form quadratic equations of which the following are roots: 
eee 

Sotution : When the right-hand member of the equation to be formed 

is 0, the left-hand member has factors x — 3 and x — 3. Hence 
fo 3 
at Petre ee er os 
(v — 3)( — 9) =2? -ar +5 = 0 

is a quadratic equation with roots 3 and 3. Since we may multiply both 
members of the equation by 2 we may write the equation freed from frac- 


tional coefficients in the form 
2a" = 12. +5 =, OU. 


2. 2, >. Ter b= Bo 1 -eNeas 

3. 3, 8. ce mon 

4. —4, —3. GD hia Fae: 

5. +/3,.-vV3> 10, OTN oe Aes 
6. a, 6. . 11. SS ee expe ae 


7 m 


12. Form an equation whose roots are 2 and * such that the 


coefficient of x? is 6. | 
2 

3 such that the 
coefficient of x is 5; such that the known term is 7. 


13. Form equations whose roots are 2 and 


82. Nature of the roots of a quadratic. In Art. 79, we 
found the two roots of the quadratic equation, 
ax? + br +c =0, 


| 35 b? — 4a 
a 
mp 4/07 — 4ac 
and Lo = 
2a 


In case a, b, c are real numbers, the numerical character of 
these roots depends upon the number b? — 4ac under the radical 


132 QUADRATIC EQUATIONS [Cuap. IX, 


sign. An examination of x and 2 leads at once to the follow- 
ing conclusions : | 
(1) If 6? — 4ac>0, the roots are real and unequal. 
Thus, in the equation 
ov? — 1llz +10 =0, 
we find b? — 4ac = 112 -4-3.-10=1>0, hence the roots of this equa- 
tion are real and unequal. Upon solution they are found to be 2 and >. 
(2) If 6? — 4ac<0, the roots are imaginary and unequal. 
Thus, in the equation 
x? — 62 +58 =0, 


we find b? — 4ac = 6? — 4- 58 = —196<0, hence the roots of this equa- 
tion are imaginary and unequal. Upon solution they are found to be 
3 — 7i and 3 + 7. 


(3) If b? — 4ac = 0, the roots are real and equal. 
Thus, in the equation 
4a? + 4 +1 =0, 


we find b? — 4ac = 42 -4- 4-1 =0, and the roots are real and equal. 


Upon solution they are found to be —2, —3. 


It should be observed that if the coefficients are real and one 
root is imaginary, then both roots are imaginary. | 
The expression b? — 4ac is called the discriminant of the 
equation 
ax? + bx +c = 0. 
If we add together the two roots of the equation 


ax? + bx +c = 0, 


we have 
-b. ++/b2 tue . — b —\7b eae b 
ay +t = —— 5 + — = - 
2a 2a a 

Hence: 


I. The sum of the roots of a quadratic equation in x is equal to 
the coefficient of «2 with its sign changed, divided by the coefficient 
Of a7. 


Art. 82] NATURE OF ROOTS 133 


Thus in the equation 
dav? — 444+ 5 =0, 


we know immediately that the sum of the two roots is . 


If we multiply the two roots together, we have 


—b ++/b? — 4ac\f —b a ie! C 
ae a SY rad 


Hence: 


Il. The product of the roots of a quadratic equation vn x vs equal 
to the known term divided by the coefficient of x. 


Thus, in the equation 3x? — 4x + 5 =0, 


we know immediately that the product of the two roots is bs 


3 
EXERCISES 
Determine the nature of the roots of the following equations : 
Trtig? — 237+ 2 = 0. 3. 27? + 6x + 199 = O. 
2. 27— 2c + 2 = 0. 4. 4977+ 147 +1 =0. 


& ytyt+1 =. 
Determine the real values of k so that the roots of the follow- 
ing equations may be equal : 
6. w+r+k =0. 
SotutTion: In order that the roots of this equation may be equal we 


must have 6? — 4ac = 1 —4k =0. Hence k is }. Substituting this value 
in the equation we have 


wP+e+s = (© +7)? =0. 
7 v@tke+1=0. 9. v2 +3kr+k+7 =0. 
8. 2a? 4+ 824 +k = 0. 10. kx? —- 4x4 +3 =0. 
Determine by inspection the sum and the product of the 
roots of the following equations : 
di ae dae 1 = OQ. 13. 2+ 5x — 2a? = 0. 
12. y+ 8y — 11 =0. 14. 2? — 274 + 3k = 0. 
15. 2 +pr+q=Q0. 


134 QUADRATIC EQUATIONS [Cuap. IX. 


Determine the value of k in the following equations : 


16. 2? + kx + 3 = 0, where one root is. 2. 


Sotution : Let 2; be the second root. The product of the roots of this 
equation is 3. 








Hence, Pies 1 = 13; or 2; -3. 
aet er ta AH a i The sum of the roots is —k. 
-—— Be iad Hence, 2 +5 ee —k 
eet pea per i: 
| HH are and k=—>5 


) | | 17. x? + 5¢ 4+ 2k = UF 
aes! | Se E where one root is 2. 

| | | i 18. 227 + ke + 14 = O, 
PCE TW, ae eae where the difference between 
| ° | Y the roots is 9. 
HEEH | b | LI 19. 2x? — 6kx + 9k = 0, 


Se pana me eS where one root is 6 times 
| \ / See ee the other. 
U HH 20. 627 + kz — 7Fo=10; 


where the quotient of the 








/ eo ee n ° =F 
0 ig Ripert trie nae two roots is——- 
Fig. 36 2 


83. Graph of the quadratic function. To graph the quad- 
ratic function x? + « — 2, compute a table as follows: 





























r= Fe sede kee ae hee 
e+2—-2=| 63] 4] 18| 0|—13|—2|—22|/—2|—12/0/92/4 
- Plotting the points (—3%, 63), (—3, 4), - - - - from the table and 


drawing a smooth curve through them we have the curve in 
Fig. 36. The graph crosses the X-axis at —2 and 1, hence for 
these values of x the function x? + 2 — 2is zero. In other words, 
—2 and 1 are the solutions of the equation 


r+ 2 — 2'= 0, 


Arr. 83] GRAPHS OF QUADRATIC FUNCTIONS 135 


They are represented graphically by the abscissas of the points 
where the graph crosses the X-axis. 

It can be shown, if a is positive and different from zero, that 
the graph of the function az? + bx +c has 
the same general characteristics as the curve 
in Fig. 37. This curve is called a parabola. 
The real roots of the equation 


az” + bz +c = 0 


are given by the abscissas of the points 
where the curve crosses the X-axis. If the 
curve has no point in common with the 
X-axis, then the roots of the equation are 
imaginary. If the curve touches the X-axis, both roots of the. 
equation are real and equal. These three cases are shown in 

Y Fig. 38, where the graphs 
of 7? +2—-—2, +244, 
x? + a+ 2 are shown. 


Kiev 3/ 


EXERCISES 
Construct the graphs of 
the functions in the fol- 
lowing equations, and by 
measuring determine the 
roots if they are real : 
Lea to), 
a? 22 —3 =0. 
x?—2x4+5=0. 
7—-2744+1=0. 
2a? + 3x —2 =0. 
2x? + 34 +2 = 0. 
Av? — 122 +9 =0. 
8. 1+ 27-2? = 0. | 9) 2? 5 Sea 
Teor — a2 = (). 

















~ | Scale, 2 spaces=1 Unit 
‘77 


Fic. 38 


TS eS ee 
8 
b 














136 QUADRATIC EQUATIONS [Cuap. IX. 
MISCELLANEOUS EXERCISES AND PROBLEMS 
Solve: | 
1. a? — 97 + 14 = 0. 3. 1277 —2z —~1=0) 
2.) 2? bor + li 0: 4. 77+ 54-1 =0. 
5. 5x2? + 247 — 5 = 0. 
6 («+1)?4+3¢4+1) +2 =0. 
2 3 
3 11. ——— 4+ —— =], 
ie BD ty ters 243 Lae: 
x 
. n m 
8. A 9. $2: A ——— =], 
x+2 — + —+n 
x 
1 Lema x—-a 2%a—2z)—b 
rosette Oi: tet ba 
iad: Wome “2, 
10. = a. ££) °a ae 
CF: see — + — 
a. ££ 
15. Find all of the roots of the equation x? — 9x = 0. 


Solve the following equations: 
16. x? — 7.35% + 2.45 = 0. 


Wy 


277 + O27 — 8.04 = 0. 


What are! the values of & if the following equations have 
equal roots? 

18. 2? — 2ka + 3k? = 9. 

19. kx? + 2kx — 32 42 = 0. 


20. 


Find by the graphical method the approximate values of 


the roots of the equation x? — 4x — 13 = 0. 


Find the quadratic equations whose roots are : 


21. 
22. 
23. 


Tata's 
(a — b), (a + 5b). 
n(1 +n), n(1 — 7). 


Art. 83] | PROBLEMS 137 
24. For what value of k will the equation 


a 2 ke 1. 
ied Oty ea | 





have roots arithmetically equal but opposite in sign ? 

25. Find two numbers whose sum is 12 and the sum of whose 
squares is 74. 

26. What numbers differing by 9 have a product 198 ? 

27. Find two consecutive integers whose product is 506. 

28. Find two consecutive even integers whose product is 
288. 

29. Separate 42 into two parts such that the first is the 
square of the second. 

30. Separate the number a into two parts such that the first: 
is the square of the second. 

31. One leg of a right triangle is twice the length of the 
other. The number of feet in the perimeter of the triangle is 
the same as the number of square feet in the area. Find the 
sides of the triangle. 

32. One leg of a right triangle is n times the length of the 
other. The measure of the perimeter in feet is the same as 
the measure of the area in square feet. Find the sides of the 
triangle. 

33. One leg of a right triangle is 5 feet and the hypotenuse 
is 5 feet longer than one half the other side. What is the area 
of the triangle ? 

34. The length of one leg of a right triangle is a and the 
hypotenuse is a feet longer than one half the other side. What 
is the area of the triangle ? 

35. Find the radius of a circle such that the number of feet 
in its circumference equals the number of square feet in its 
area. | 

36. By increasing the radius of a sphere by 1 inch, its 
volume is increased 40 cubic inches. Find the radius of the 
original sphere. (See 17, p. 64.) 


138 QUADRATIC EQUATIONS [Cuap. IX. 


37. The area of a triangle is 16 square feet and the base 
is 12 feet more than twice the altitude. Find the base and 
altitude. 

38. The area of a triangle is b square feet and the base is 
n feet more than twice the altitude. Find the base and altitude. 

39. Show that the equation x? — ax — b? = 0 has one posi- 
tive and one negative root. 

40. Around a rectangular flower bed which is 3 yards by 4 
yards, there extends a border of turf which is everywhere of 
equal breadth and whose area is 10 times the area of the bed. 
How wide is the border ? 

41. Graph on the same sheet of codrdinate paper the function 


a7 +4+6¢ 


where c takes on the values —3, 0, 1, 10. What effect does 
changing the constant term in a quadratic function have on — 
the graph ? 

42. Graph on the same sheet the function 


ax? +241 


where a takes on the values 10, 1, 3, 75. Decreasing the coeffi- — 
cient of x? has what effect on the graph ? 

43. The edges of a cube are each increased in length 1 
inch. It is found that the volume is thereby increased 331 
cubic inches. What was the length of the edge of the cube? 

44. A cistern is filled by two pipes in 40 minutes and the 
larger pipe can fill it in one hour less time than the smaller one. 
In how many minutes can the smaller pipe fill the cistern ? 

45. A rectangular piece of tin is three times as long as it is 
' wide. From each corner a 3-inch square is cut out and the 
ends turned up so as to make a box whose cubic contents are 
312 cubic inches. What were the dimensions of the piece 
of tin ? 

46. A man sold some railroad shares for $4050. <A few days 
later, the price of the shares having dropped $6 per share, he 


Art. 83] PROBLEMS 139 


used the money to buy 4 more shares than he had sold. How 
many shares had the man in the first place ? 

47. The sum of the numerator and denominator of a frac- 
tion is 20. If the numerator is decreased by 3. and the denomi- 
nator increased by 3, the resulting fraction is half a great as 
the original. What is the fraction ? 

48. If $4000 amounts to $4410 when put at compound 
interest for two years, interest being compounded annually, 
what is the rate of interest ? 

49. A crucible contains 250 pounds of melted lead. A cer- 
tain quantity of the lead is dipped out and replaced by the 
same amount of melted tin. From the mixture the same 
amount is removed and again replaced by the tin. The final 
' mixture contains 16 parts of lead to 9 parts of tin. How much 
lead was dipped out each time ? 

50. A and B together can address 180 envelopes in an hour. 
When each works separately at 180 envelopes, A finishes 27 
minutes before B. How many envelopes can each address in 
an hour? 

51. In joining together two steel boiler plates with a single 
row of rivets, the distance p between the centers of the rivets 
is given by the formula 

d? 
p= 0.56-- + d, 
where ¢ is the thickness of the plate and d the diameter of the 
rivet holes. In a boiler the rivets are to be placed 2 inches 
apart. If the thickness of the plate is ¢ inch, what is the diame- 
ter of the rivet holes to the nearest sixteenth of an inch ? 

52. Work Problem 51 for rivets 1 inch apart. 

53. A company of boy scouts engaged to carry a message 84 
miles by running in relays. Six of the company were unable 
to run and it was found that each of the others would have to 
run one third of a mile more than had been expected. How 


many boys were in the company ? 


140 QUADRATIC EQUATIONS [Cuap. IX. 


54. The density of dilute sulphuric acid when 100 grams 
contain s grams of the pure acid, is given by solving the 
equation, | 

1.35D? + 112D — 111 —s =0, 


for D. If s = 3.5, find D to two significant figures. 
55. The temperature drop below 212°, at which water boils 
at elevation h above sea level, is obtained by solving 
t??+ 517th =0 


for t. At what temperature does water boil at an elevation of 
10,000 feet ? 


CHAPTER X 
SYSTEMS OF EQUATIONS INVOLVING QUADRATICS 


84. Introduction. A quadratic equation in two unknowns 

x and y is an equation that may be written in the form 

ax? + by? + cry +dx+ey+f=0, 
where a, 6, and c are not all equal to zero. Thus, x? + y? = 10, 
y?+x2=5, zy =1, and zy+ 2x +y =0 are quadratic equa- 
_ tions in two unknowns. 

We shall be concerned in this chapter only with quadratic 
equations of certain special forms. The solution of equations 
of these special forms will be useful in solving some interesting 
problems. 

85. One equation linear and one quadratic. Any system of 
equations in two unknowns in which one equation is linear and 
the other is quadratic, can be solved by a method of substi- 
tution. 

Example 1. folve the system 


x + y? = 25, (1) 
by = 1. (2) 
SoLuTION : Solving (2) fory, y=ax-1. (3) 
Substituting « — 1 for y in (1), 
x7 + (¢ — 1)? = 25. (4) 
From ( 4) 2x? —2x — 24 = 0, 
or BF mn AZ =I, (5) 


Solving (5) by the formula 
} 1+v1 +48 
a cae Te ts 
Z, 
=4or —3. 
Substituting 4 for x in (2), weobtain y =3. 


Substituting —3 for x in (2), we obtain y = —4. 

This gives } eke and See for the solutions. 
y= ys = 4 

Check these solutions by substitution in (1) and (2). 


142 EQUATIONS INVOLVING QUADRATICS [Cuaap. X. 


Graphical meaning of the two solutions. We may plot the graphs for 
each of the equations (1) and (2). The graph of x — y = 1 is the straight 
line shown in Fig. 39, and the graph of x? + y? = 251s the circle there shown. © 
To draw the graph of (1), the student may give various values to x and cal- 

culate the corresponding values 

for y from y = £V25 — 2. 

Any point on the straight line 

(2) has coérdinates that satisfy 

equation (2). Any point on the 

circle (1) has coédrdinates that 
satisfy equation (1). The points 

(4, 3) and (—3, —4) lie on both 
x graphs, and satisfy both equa- 
tions (1) and (2). That is to 
say, each point of intersection of 
the graph of (1) with the graph 
of (2) gives a pair of numbers 
that is a solution of the system. 











ro 
co 


Unit 





Example 2. Solve the system 


xy = 2A, 1 
Fig. 39 y i: ay 0. 2 
SoLuTION : Solving (2) for y in terms of 2, 
y= 22 — 2. (3) 
Substituting 2x — 2 for y in (1), 
x (2% —2) = 24, (4) 
2x? — 22 — 24 = 0, 
a? — 7 — 12 =, 
or (x +3)(2 —4) =0, (5) 
or x = 4 or =3. 


Substituting 4 for xin (1), we have 


tee, 
Substituting —3 for x in (1) 
y= -8. 
This gives serie 2 and J” +’ for solutions as may be verified by 
ly =6 UYseiaS 


substitution in (1) and (2). 


Graphical meaning of the solutions. The graphs for equations (1) and 
(2) are shown in Fig. 40. The graph of y — 2x + 2 = 0 is the straight line, 


Arr. 85] EXAMPLES 143 


and that of zy = 24 is the curve with two branches as shown. This curve 
belongs to a class of curves called hyperbolas. The points of intersec- 
tion, P and Q, have coérdi- . Yy 

nates that are the solutions of 
the given system. All points 
on the graph of (1) have co- 
ordinates that satisfy equa- 
tion (1). All points on the 
graph of (2) have coérdinates 
that satisfy equation (2). | / 
Therefore, the points of inter- X——— ~ ee xX 
section have coérdinates that | 
satisfy both equations. 















| 
a 


Example 3. Solve the 
system 
x? + y? = 25, (1) 
z+y = 10, (2) 
and draw the graph to explain 
the fact that the solutions are 
not real. 
If the student will carry out the same method as that illustrated in 
Example 1, he will obtain for solutions 


; 
/ 
| 


Scale, 1 space=2 Units 


2-5 +> 2 
Ot Wie 

and x =5 — > 2, 
Se ee. 
emer A; 


and, y= 5. x JE, 
where 7? = —1. 


Check these solutions by sub- 
stitution in (1) and (2). 
The graph of equation (1) 
is the circle shown in Fig. 41, 
and the graph of equation (2) 
is the straight line there 
Fig, 41 shown. Itis to be noted that 
these graphs do not intersect. This fact means that there exists no pair 
of real numbers that satisfies both equations (1) and (2). 








144 EQUATIONS INVOLVING QUADRATICS [Cuap. X. 


EXERCISES AND PROBLEMS 


Solve the following systems, verify each set of roots by 
substitution in the given equations, and draw the graph for 
at least Exercises 2, 3, 5, to show the graphic interpretation of 
the results. 


j Rag era Os The Fn Bs 9. 7 — reeds 
x+y =d. ry =a?+a. 

2. «7? — y? = 48, 10. 2* + y? = 260, 
x — ly = 0. LY ae 

Pp Gre oa het MO 11. xy + 150 = 0, 
2x + 3y = ls. x—y+31=0. 

4. 5x? —3y?+7 =0, 12. 2x? — 2ry + y? = 10, 
2e+y—-—7=0. Cys TT 

5. ry + 54 = 0, 13. x—y = 4, 
rt+y+3=0. _ dx”? + Qry + 4y? = 25. 

CS ee re A BP 14. 7+y+2y = —4, 
xy +15 = 0. r—y=4., 

ie 2) 1 1. +71, 
a ide OP ha) el a ij 

8. 44 — 3y + 16 = 0, 16. 2 ee 
Dee se 
cee 47 + dy = 3. 


17. The perimeter of a rectangular athletic field is 1124 
yards, and the area is 16 acres. What are the dimensions? 

18. The difference of the two legs of a right triangle is 7, 
_ the hypotenuse is 17. Find the sides of the triangle. 

19. Find two consecutive integers the difference of whose 
squares is 31. 

20. The difference of the areas of two squares is 700 square 
feet and the difference of their perimeters is 40 feet. Find a 
side of each square. 


Arts. 86, 87] SIMULTANEOUS QUADRATICS 145 


86. Both equations quadratic. When both equations of a 
system are quadratic, the problem is often so difficult that the 
system cannot be solved by methods at present at our disposal. 
There are, however, some forms of such equations for which 
we may easily obtain solutions. In Arts. 87 and 88 we shall 
consider a few such equations. 


87. Both equations of the form aa’?+by?+c=90. If, 
instead of considering x and y as the unknowns, we consider 




















— ee ee 


Scale, 2 spaces=1 Unit 





Fig. 42 


first x2 and y? as the unknowns, the method of solution is that 
for linear equations. 


Example. Solve 9x? + 16y? = 288, (1) 
| x2 4 y? = 25. (2) 

SoLvuTIon : Solving for x? and y?, we have 
x? = 16, (3) 
y? =9. (4) 


From (3) and (4), g=4+4,y = +8. 


146 EQUATIONS INVOLVING QUADRATICS [Caap. X. 


Iorming all possible pairs, we have the four solutions 


ly =3, ly = -3, 7 aoe ly = -3. 


The equation (1) has for its locus an oval shaped figure called an 
ellipse (Fig. 42.) The equation (2) has a cirele for its locus. The four 
points of intersection represent graphically the four solutions. 


EXERCISES 


Solve the following systems of equations and give the graphical 
representation for 5, 6, 7, and 8. 


1. 9x? + 25y? = 225, 5. 9x? + 16y” = 288, 
i oh OP) SY x? + y? = 9, 

2. O08 Lge ee G0, 6. 9x? + 16y? = 288, 
7 yr Lay, e* 4p eas 

3. 49? —9y? ‘=.19, 7. 9x? + 16y? = 288, 
x? + y? = 34. x? + y? = 32. 

4. 4074+ y? = 61, 8. 9x 7+ 16y? = 288, 
2x? + 3y? = 93. x? +y? = 64. 


88. Further special methods. Many systems may be solved 
by special devices in which the aim is to find values for any two 
of the expressions, x + y, x — y, and xy, from which the values — 
of x and y may be obtained. Other devices, such as substitu- 
tion, are suggested in the exercises which follow. 


Example 1. Solve the system 


x? +2y = 12, (1) 

y? + ay =4. (2) 
SOLUTION : 
Adding (1) and (2), x? + 2ry +y? = 16. (3) 
Hence, each Oye: clera (4) 
Subtracting (2) from (1), Co dat) bart (5) 
(5) + (4) gives, —-y=+2o0r—-2. © (6) 


From (4) and (6), Zee oor —3; y = 1 or —1. 


ArT. 88 | EXERCISES 


By substitution in (1) and (2), we find that only the two pairs 


eee 


3 : 
ey i satisfy (1) and (2). 


Example 2. Solve the system 
x+y’ = 16, 
a? = OF. 
SoLuTION: Substituting 6x for y? in (1), we have 
x? + 6x = 16, 
or x? + 6x — 16 = 0. 


By formula (Art. 79), 2 = aes 


| 


If x = 2, we have y = +V/12 = + 2V3. 


147 


(1) 
(2) 


(3) 


If x = —8, we get y = +V—48 = +4i\/3, which are imaginary numbers. 


The solutions are (2 =2 “i ~=2 c= —-8 x= -8 
Peat y 2/8 Ny = a5) 


Check the results by substitution. 


EXERCISES 


Solve the: following systems: 


1.y* = 62, 6. 27 + zy = 36; 
x—y+8=0. ry + y* = 45. 

meee ey? = 8), Tee ri 10) 
x? — y? = 77. ry + y? = 6. 

3. 27+ y? = 25, 8. p?—q? =9, 
cy = 12. 4p? = 25g. 

4, 5x7 — 9y? + 121 = 0, 9. +7 = 34, 
Ty? — 3x7 — 105 = 0. Sie to; 

6. 9aee a = 50} 10. s* + st = 45, 


ry = —7. s? — st = 5. 


y = —4ivV/3. 


EQUATIONS INVOLVING QUADRATICS [Cuap. X. 


MISCELLANEOUS SIMULTANEOUS EQUATIONS INVOLVING 


148 
QUADRATICS 

Solve : 

1)? = 77 = 100, 14, 
ox + 4y = 50. 

2.2—y = 9, 15. 
xy = —6. : 

Si 97? 2-254? = 225, 16. 
Ox? — 25y? = 225. 

4. 477 = 977, ti 
4x? + Oy? = 1. 

hee rie ee) lean 8 


Hint: Subtract second from first. 


6. 


x?—xzyt+y’? = 3. 


x—y =0, 
Cy? = 

2a) — 3¢Y en 

bx + y = 3. 

LY — By = 5, 

2y + ry = 6. 

xv? +y? + 3x + dy = 28, 
xy —6 = 0. 


Hint: Multiply second by 2, add 
to first and solve for x + y. 


18. 


19. 


eaters 
xy —10 = 0. 
2(z — y)? +2(4 + y)? = 85, 


Qa —y) + 4(a + y) = 25. 


ey 12 =0; 

x? + y? — 25 = 0. 
x? — y* = 3, 
C4 = Ole 


Hint: Divide first by second. 


10. 


11. 


12. 


13. 


4x? — y? = 16, 
2 Ay = se 
r+ y = ary,’ 


etyt+ar+y? = 49. 


x? — Qey — 2y? = 32, 
x — by = 2. 
x? + 4y? = 24, 


CY nO, 


20. 


21. 


e+y+art+y =2, 
xy —2=0. 

ty = 2(r + y), 

x? + 2ry + y? = 1. 
22? + 5y? = 125, 


5a + 2y = 10. 
2x* — zy = 10, 
Ot" = Ate 


Hint: Multiply first by 11, second 
by 10, and subtract to. eliminate the 
constant terms. Factor the resulting 
expression into 


22. 


23. 


24. 


(x + 2y) (8a — 5y) =0. 


x? + xy + 2y? = 44, 
2x? — ay + y? = 16. 


x? +y? = 16, 
a? y” 

Og. (eu 
x? + y? = 25, 
ox — 2y = 6 


Art. 88] EXERCISES AND PROBLEMS 149 


25. 27 + 4y? = 17, 31. 6x? + dry — by? = 0, 
wipe 2: 2a? — y? + 5x = 9. 
26. v?+y’?=9, 32. 274+ sry + y? = 41, 
ot — 2. ety tety = 32. 
aT ee OY 4, 33. 2? + dry + 2y? = 3, 
Po y= 18, Qr? + y? = 6. 
2 va le 
28. 3x” + dxy + 2y” = 8, 34. (x + y)2—3(x + y) = 10, 
x? — xy — 4y? = 2. g? — y? = 


Hint: Let y = vx. Substitute in 
both equations and solve for v and z. 


= 5, 

S0 ety 10, 

UY ie Te 
36. 2? +y?+x—-—y-—12=0, 


tomes oY ced OF 


29. x7 + 32y — 2y? = 2, 
2x? — dry + by? = 3. 


30. x77 + zy — by”? = 0, 


a? + 4x + 3y = 69. 37. x? + 2ry + y? + 2x + 2y 
Hint: Solve the first equation . te bes, : 
for x in terms of y, and substitute v* — dry + y* — 2x + 2y 
in second. +1=0. 


38. The sum of two numbers is 12 and their product is 35. 
Find the numbers. 

39. The sum of the squares of two numbers is 100, and their 
difference is 2. Find the numbers. 

40. Find two numbers whose product is 150 and whose 

err 
quotient 1s 5 

41. The area of a rectangle is 120 square yards and the 
diagonal is 17 yards. Find the sides. 

42. A rope 48 feet long exactly surrounds an enclosure in 
the form of a right triangle of hypotenuse 20 feet. Find the 
other sides of the enclosure. 

43. A loop of twine 34 inches long is to be put around four 
pegs set in the form of a rectangle of area 60 square inches. 
Find the sides of the rectangle. 


150 EQUATIONS INVOLVING QUADRATICS [Cuap. X. 


44. A farmer raised broom corn and pressed 6120 pounds 
into bales. If he had made each bale 20 pounds heavier, he 
would have had one bale less. How many bales did he press 
and what was the weight of each ? 

45. After a mowing machine had made the circuit of a 
10-acre rectangular field 33 times, cutting a swath 5 feet wide 
each time, 23 acres of grass were still standing. Find the 
dimensions of the field. 

46. A farmer bought 5 cows and 30 sheep for $525. He 
bought one more sheep for $25 than he bought cows for $300. 
Find the price of each per head. , 

47. A field contains 9 acres. If its length were decreased by 
20 rods and its width by 4 rods, its area would be less by 
4 acres. Find the length and width. 

48. If the length of a diagonal of a rectangular field of 30 
acres is 100 rods, how many rods of fence will be required to 
enclose the field ? 

49. It took a number of men as many days to pave a street 
as there were men, but had there been five more workmen em- 
ployed, the work would have been done 4 days sooner. How 
many men were employed ? 

50. The sum of the squares of two consecutive integers is 
1301. Find the numbers. 

51. A rectangular piece of tin containing 400 square inches 
is made into an open box, containing 384 cubic inches, by cut- 
ting out a 6-inch square from each corner of the tin and then 
folding up the sides. Find the dimensions of the original’ piece 
of tin. 

52. If the product of two numbers is increased by their sum, 
_ the result is 79. If their product is diminished by their sum, 
the result is 47. Find the numbers. 

53. For two numbers a and 6 it is found that the expressions 

a (a — b), a? — b 
give the same number. What is this number? 


Art. 88] PROBLEMS 151 


54. The area of a field is 367,500 square feet. The length of 
the diagonal is 875 feet. What are the dimensions ? 

55. The difference of the squares of two numbers is divided 
by the smaller number. The quotient is 4, and the remainder 
is 4. If the difference of the squares of the numbers is divided 
by the greater number the quotient is 3 and the remainder 3. 
What are the numbers? 

56. A certain floor having an area of 96 square feet can be 
covered with 432 rectangular tiles. If the workman use a tile 
one inch longer but one inch less in width, it takes 512 tiles 
to cover the floor. What are the dimensions of the two sizes 
of tiles? | 

57. A club of boys bought a motor boat for $192. Four boys 
failed to pay their share as agreed, so each of the others was 
compelled to pay $4 more than he had promised. How many 
boys were in the club? 

58. A father divided $1000 between his two sons and kept 
it for them at simple interest until called for. At the end of 
24 years, one son called for all the money due him and received 
$632.50. At the end of 3 years the other son received $531 as 
his share. How was the money originally divided and what 
interest did the father pay ? 

59. A few days after the outbreak of the European war the 
cost of 100 pounds of sugar in the United States was $3.10 
more than it had been just before the outbreak. For $330 
a grocer received 3100 pounds less sugar after the outbreak 
than he would have received before for the same amount. 
What was the price before and after the outbreak of the war ? 


CHAPTER XI 
EXPONENTS, RADICALS, AND ROOTS 


89. Positive integral exponents. The expression @”, when 
n 1s a positive integer, means the product a:-a-a... ton 


factors. 
In operating with positive integral powers we make use of 


the following laws: 











1k OO = 1 
Example. os ON a 
1H fs a” +a" =a”, wherem > n. 
Example. 47 + 43 = 44, 
ITI. (7) a 
Example. (O4)* sess, 
IV. (a°bic" is -)n = arrpirmen... 
Example. (52a5c7)4 = 58q!2¢28, 
Vv. (=) a le: 
a” a”"P 

Example. (Fe) = oo 

EXERCISES 


Perform the indicated operations and simplify the results 
- when possible. Fractions should be reduced to lowest terms. 


2 a’ 
GS aN Ee Te tare 5. hea 
a 
a’ het 


2. ath? « aids, 1g aibadle 6. = 


Art. 89] EXERCISES 153 




















6min®p? — ay : 

; . . (=—])- Lies) 
i 2mn*p a (= ee 
3 12a*bc 1SapUe a 182 (ee, 
E Sab*c? 14. (—3)? - (—3)8. 19. (rrs)4, 
Be on0"C)”. azt4h§ 

10. (—3m2n’)%. 15. nebo PAVE Chal acl 
4ab gergntt ainb™ 
wie (=) 16. aALat PA at 
qzt2 (24 ei y')? ys _ ys 
22. qaz-2 , 26. ee ae iets ; hey Re ys)? 
(a <3 b)3 6aq22-8y b3y—-5 
23. RD 27. (2az2u by-9)3 
6a\?2 /b\8 b \2 (a — 1)3” 
 (F) (23) (se) ae 
a2\" c2\n xy n (6a2x3y)? 
Be) (pe) (or) aw 
CR 
31. Find the value of 1 +2 ae : gto : + ;. for x.= 10. 


32. 


33. 
34. 


35. 





exis otc 4h 4442 +5 ; for x = 10. 


Find the value of 22% + x +3 . 2 + 7a 2 ; for x = 10. 
Find the value of each of fi ae sums for z = 4: 


(a) 1+2. 

(b) l+2a2+4 2%. 

(c) L+aua+a2? 4+ 2%. 

(7) l+a+274+ 2° 4+ 24, 


Find the sum of 2, of 3, of 4, and of 5 terms of 


ee fore = 10; 
So a 


154 EXPONENTS, RADICALS, AND ROOTS [Cuap. Ub 


36. Find the value of 5é + 6#2+9t+3 for t=10. Also 
Brdiee sae enees 
eet ae | 
37. Write in terms of powers of ¢ = 10, the numbers 8648, 
48609, and 100101. 


38. Expressed in terms of powers of ¢ = 10, 


35.762 = 31+ 5 +i +4 Ba 
Write in terms of powers of t = 10, the numbers 3.1416, 1.4142, 
and .06041. 


90. Extension of the meaning of an exponent. Thus far we 
have dealt entirely with positive integral exponents. No 
meaning has thus far been given to a” if n is zero, fractional, 
or negative. It is found useful to give a meaning to such 
expressions as a°, 8?, and 2°. It is convenient to define them 
so that the laws for positive integral exponents hold for these 
new kinds of exponents. Such definitions will now be given. 


91. Meaning of a. In order that Law I, Art. 89, may hold 
for an exponent zero, it is necessary that 
CE. (UT el Ch at eset 
or a =)1) 11.a = 0 
That is, for all values of a, except 0, 


ao =1, 


92. Meaning of @’. Assuming that Law I, Art. 89, is to 
hold for fractional exponents, . 
at - at. at = aititt =a, 
and a? is one ‘of the three equal factors whose product is a. 
But this is what has been defined as the cube root of a. Hence 
we say that he 
a= Va. 


Arts. 92, 93, 94] NEGATIVE EXPONENTS 155 


In general, if g is a positive integer, 
eer ne... to g factors = av eta tT ** a terms 
= 4, 
and a’ 1s one of the q equal factors of a, or a qth root of a. 
That is, ai = Va 
Examples. gi = V8 =2, 
50? = 50 = 5V2. 


QIK 


93. Meaning of nee Let p and q be positive integers. Then 
applying Law I, Art. 89, we have 


Pp 


, +ott+...topterms _q@ 
) 


at?-a’-at...to p factors = a! 
p 
and a% means the pth power of the gth root of a. That is, 
| P &y ae 
aa = (s/a)? aa Va". 
Examples. 273 = (0/27)? = 9, 
9? = (V/9)5 = 243. 


We have now found meanings for zero and _ fractional 
exponents. 


94. Meaning of a” when mm is negative. Let m = —n, where 
n is a positive number. Applying Law I, Art. 89, we have 


a-a*=ar"=@ =1. 
i ee 
Hence, Cech cer. Urdu 0: 


1 


a™ - 





and | a” = 


Thus, a” and a~” are reciprocals of each other. 


Historical note on exponents. It appears that Nicole Oresme, a bishop 
of Normandy, in the fourteenth century conceived the idea of fractional 
exponents, and gave some rules for operating with them. But his nota- 
tion was very different from ours. To Simon Stevin of Bruges in Belgium 
(1548-1620) belongs the honor of inventing our present method of de- 


156 EXPONENTS, RADICALS, AND ROOTS  [Cuap XI. 


Therefore in a fraction, any factor of the numerator may be 
changed to the denominator, or any factor of the denominator 
to the numerator, if the sign of the exponent of the factor is ° 


changed. 
27a-*b = 22-5 b 


5243 2a 





Example. 


We have now found meanings for zero, fractional, and 
negative exponents. 


EXERCISES 
Give in simplest form the values of the following : 
Depots oO walt es eee 
2.502 e14e ee 
Sree oars Dy 
| as (ae 
aig ey 16. (5y5)?. 
5 17. (3)° ie 
5. 5-1 
187-1250 
| 1\-2 ; 
Die! (Gy 19. 648. 
. ~20. 8-1(3)° - 363. 
ee (3) 
en 
—8. (x + 2)°. 21. 14 34 
9; (=o) oe. —2 «2 
oo, 7am 
10. 16: lis 
11. 83. —23. (394,)9 + 84 — Gj. 
12.0 2ote Ace 24, 2-37-67. 


25. 8 - 273 — 27 - 83 + 8-3 - 27-4 — 8 . 273. 
26, 2-5 — 5% + (§)?— 
noting powers. It ‘remained, however, for John Wallis (1616-1703), an 


Englishman, to show the advantages of fractional and negative exponents 
to such an extent as to assure for them a permanent place in algebra. _ 


Art. 94] EXERCISES ; 157 





yo) ieee ES ie 29. 3 - 32% — 3 - 32! 
i> > i 
2 pe ie . i ten Je (—1)4 
fee 2 + 374), at ( 14° — Qu ): 


Change the following into identical expressions without zero 
or negative exponents; and simplify when possible: 











Pain O27". a‘ 
45. ees 
32. (x? + y”)°. a*+a 
oe Gets te LO, 
SotutTion: By Art. 94, : 47. (a —b)- (a2? — Bb). 
Ba2e-2y-8 = 5a? - 1 i. = 5a? 
mys oy =48. (1 a) (a a) 
0 
ok. 49. 4x fy + 2). 
Borg 2 2. ie aie 
3-2 a?bm 
She a on 
51. . 
m2n—4 aib—ic8 
St. ns 52. (22 + y)* (2? + y)!. 
SoLuTION: By Art. 94, 63. a-! + 2a? + 3a-3 
mn-* min? m4 , ; 
m ns ni n- 54 Z 
38 12ab-3 a7) + 2a-* +. 3a-° 
8a 55. a! + (2a) + (3a). 
3 
39. 32" — —- bree ee. 
% "72 y? 
B02 4a) (—4).°. j 
Al. aty t: ~ 57. ae 
Pot. 
7 1 
ke a ‘be? =58. 2a% — dq-2 
1 | 
———: 2 =8 
“er 2 (y-" 
—2 —2 , +P © 
a 


gy? 60. 8° + 81 + 8 — 84. 


158 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. 


Change the following into identical expressions without — 
fractional or negative exponents: 


Slee. 
62. ai. 
63. 8 - 53, 
64. 
65. 24. 


n 
66. ym. 


67. te 





68. 





69. e€ 
TO 


71. 
72. 
73. 
74. 
75. 
76. 
ads 
78. 
79. 
80. 
81. 


(m + n)?. 


(8xy°)*. 
amen’, 
12a7?. 
a*b-*c8, 


(72 + 4)—4, 


(k-) 4 1-8), 


Tx Ty!, 
[(a))-1 


82. 


83.0 


84. 


85. 





86. yi+y?+y. 
- y? + y? + y?. 
3 gel ete gel, 


. 8a® \3 
89. (E) 


90. 


fo) 


Os 
3 225 


Change the following into identical expressions without 
radical signs or negative exponents: 


91. 
92. 
93. 
94. 


SOLUTION : Va*b'ct = (ab5c!)2, by 
definition of a fractional exponent, 
= a*bc2, by Law IV, Art. 89. 


95. Wa: Vb. 
96. v/a"bce, 
97. v/a%b’c. 
98. Wa + b. 
99. v/(a + b)*. 
100. v/a® + B®. 


101. 


102. 


103. 
104. 
105. 
106. 


107. 
108. 


109. 
110. 


5 gly 

mM 5930 4 
\/ ab?c3 | 

rye! 
Vath » Wx. 
SOO, 
Va 5, 
Vat. 
v/16a-8c-3. 
V8.8, 
V/36 + 64. 
V9 5. 





Arts. 94, 95] IRRATIONAL NUMBERS 159 


Perform the indicated operations. By the use of nega- 


tive exponents if necessary, leave the results in forms without 
denominators. 


oii. ~ O°. 123. (=). 

a2\3 
tiered o*. 124. (Ss) 
Plame f= x, *y?. 125. (—a*)-?. 
a4, 7. 126. (—m')-8, 
TAD yt 27°, 127. (GE): 
aoecae” + at. 128. m* + m-*. 
117. ki - ki. 129, rt}. 72, 
Pipeie ~ 73. 130. (=). 
119. (2a°b-%) + (a-%b?). 131. (rH)tH, 
120. (a?)*. TS 2H te ON, 
121. len). LS be (rot Reema 
122. (a-")~2, 134. (Qh . Qn-1)3, 


135. (a')5. 
Be ty fey) (st — a tyd + 41). 
hy (m-? + n~?) (on? —m? n-4 ve ia) 
n Dp 
138. (a ™ + ba)’, 
139; (oa? + bx-! + c)?. 


Ue Tale 
95. Rational and irrational numbers. A rational number 
is defined (Art. 67) as one that can be expressed as the quotient 


160 EXPONENTS, RADICALS, AND ROOTS  (Cuap. XI. 


of two integers. An irrational number is a real number that 
cannot be thus expressed. 


Thus, 16, 4, and 12 are rational numbers; V2,* 72, V3, V5,.1 + V5, 
and 9? are irrational numbers. 


Any irrational number can be enclosed between two rational 
numbers that differ from one another by as small a number as 
we please. 


Thus, we may write, 
1<V2<2. 
14<V/2<1.5. 
-141<V2< 42: 


Hither of the two sequences of numbers in the two outer 


1 columns determines 2 in the same way that .3, .33, .333, 
..., determine }. 
1 As a geometrical illustration of an irrational number we 


Fic. 43. may take the diagonal of a square whose side is 1. 


96. Radicals. An indicated root of a number is called a 
radical. 


Thus, V3, V64, wh </7, and Va + 6 are radicals. 
The radicand is the number of which the root is to be taken. 
* To show that /2 cannot be expressed as the quotient of two integers, — 


suppose it is possible that 
v2 =<, 


™m. ; Swi yaaene 
where pe a rational fraction in its lowest terms. At least one of the num- 


bers m or 1 is odd. Clearing of fractions and squaring both sides, we get 
ony? = 7, 

From this equation, we see that m? is an even number. Hence m is an 
even number. If m is even, m? contains the factor 4. Hence n? is an even 
number, and v is itself even. This is contrary to hypothesis that — is a 

n 


fraction in its lowest terms. 
This proof is found in Euclid (about 300 3.c.), and is supposed to be 
due to a much earlier mathematician than Euclid. 


Arts. 96, 97, 98] SURDS 161 


The number that tells what root is to be taken is called the 
index of the root. 


Thus in V3, 61/19, and 8 ~/z + y, the radicands are 3, 19, and x + y, 
and the indices are 2, 4, and 3, respectively. 


The order of a radical is given by the index of the root. 
Thus V5 is of the second order, 85 is of the third order, and so on. 


97. Surds. A surd is an irrational number that is the root 
of a rational number. 


Thus V2, 0/19, ¥/45 are surds. But v/8 is not a surd, since W/8 = 2, 
a rational number; and V 2 + V3 is not a surd because 2 + V3 is not a 


rational number. 


A surd of the second order is called a quadratic surd. 


EXERCISES 

1. Give an example (a) of a rational number ; (b) an irra- 
tional number ; (c) a radical; (d) a surd; (e) an index; (f) 
a radicand. | 

2. Give an example of a radical that is not a surd. 

3. Name the order of each of the following: V8, Wa’, 
Va + y, 

4. Tell which of the following are rational: 4, - /49, V9, 
3, 333, 1 +/5. Tell which are irrational. 

6. Tell which of the following are surds: 9, W/9, V5, 
1+V5. 

6. Give an example of a surd of the second order ; of the 
third order ; of the fifth order. 


98. Introducing a factor under the radical sign. The 
coefficient of a radical may be introduced under the radical 


162 EXPONENTS, RADICALS, AND ROOTS [Cuap. XI. 


sign as a factor of the radicand by raising such coefficient to 
a power equal to the index of the root. 


Example 1. 7/2 =V7?° V2 =V7?-2 = V8. 
Example 2. 2a2bWabx = W8a°b?Wabx = V8a'b'x. 


EXERCISES 


In the following introduce the coefficients under the radical 
sign : 


1. 573. 8. —5mnw/3m?. 

224 2. 9. 3/6. 

3. 60/7. 10. 2v/25am. 

4. 3x75. .. dL 2 7a 

5. 10/01. 12. (l+2)V1 +42. 

6. av/b. 13. (a + b)x3. 

7. 2a2yw x. 14. (m —n)Vm +n. | 
15. Va +0. 


99. Simplification of radicals. For purposes of computation 
it is convenient to be able to change the form of radicals. 
In the case of square roots, we have, that 


Vab = (ab) = at bk = Va Vb. 


- By Law IV, Art. 89, this principle can be extended. By 
this law, | | 


1 
(a"b™)k = akb 


am 
k 
) 


that is, Var” = Varrx/b, 
This principle is useful in simplifying radicals. 


Thus, — Vabs? = Va® Wx? = a2v/2?. 


Art. 100] RADICAL IN SIMPLEST FORM 163 


100. Simplest form of a radical. A radical is said to be 
in its simplest form if the radicand — 

(1) Is integral. 

(2) Contains no rational factor raised to a power whose 
exponent equals the index of the radical. 

(3) Is not a power whose exponent has a common factor with 
the index of the radical. 


Thus, V3, ~/a2b®, and ~/4ab! are not in their simplest forms, because 
they may be written as $2, bvW/a’b?, and ~/2ab? respectively. 


EXERCISES 


1. Tell which of the following are not in their simplest 
| oem8 and give reasons : 


6’ 25/50, Vv abe, 





, aor 16m2n4, Viz Vv ab? — at. 


Reduce to simplest form : 





2. V7. VN Se 
AE Me = /25-5 = pee a 

3. 18. 13. 24/4 

4. VBI. 14. 5v/2. 

Bia ON. 15. 8v/8. 

Gy 8A/72. 16. v/a’. 

7. 7/147. 17. ~/4a°b*c. 

8. 81. 18. v/16x5y'. 

Deis. 19. 3y/ 24n'yeh 

10. V5 | 20. ax? + abzt. 

Shea 21. V/m® + min. 


Pon 


3-7 . = Vor «18 


22. Wi (a + b)4. 


164. EXPONENTS, 


23. [ (x? — 


24. 


25. 


26. 


27. 


28. 


29. 


101. Addition and subtraction of radicals. 


RADICALS, 


eile 


(m2n3 + 6m3n4)?. 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


AND ROOTS [Cuap. 


1 1\34 
at 68) 
_ m4 
(+3) 

1 
(aaa 
Vary grr, 
BWA fel 
v/ 21623 y%29. 


(49a?"b8c*)?. 


>i 


Two radicals 


which, when reduced to their simplest forms, have the same 
order and the same radicand, are said to be similar. 

Thus, 5\/2 and —2v/2 are similar; so also are 7/81 and 2bv/3a' since 
4/81 = 3/3 and 2bW3a? = 2abw/3. 


On the other hand, V2 and V3 are dissimilar ; so are V2 and w/2. 


The algebraic sum of similar radicals equals the common 
radical factor multiplied by the sum of its coefficients. 


Example 1. 
Example 2. 


Example 3. 


3Va@-a+2 





bo 


8/2 +3/2 — 4/2 = (8 +3 —40 eee 
/75 +3712 = 5727 = 25 See ee 
5V3 + 6V3 -— 15V3 


= Aaa oe 
ae 2 
1/4 2aba\ _ 3b 25a? 


— 3bvV/25a‘b2a 


= 5va + eva — 15a*b?V/a 


( 


a 
2 


2b? 


2a es 242 
te 15a2b 


va 


3 -5V9-3 


ArT. 101. ] EXERCISES 


EXERCISES 


Collect terms and simplify : 


le 


Ss 
a 


ae 
bo 


a 
is) 


14. 


15. 


16. 
1G 


Re ee eek ere oe ee 


38V2 + 8/2. 

V3 — 2/3 a L173: 

/20 + 8/45 — V5. 

3/28 — 63 + 4/175. 
W/81 + 5W/24 — W375. 

16 + 9/250 — V3. 

6\/§ — /24 — V2 + 8V6. 
Va + 6Va + 2Va. 

3/3 + 4V/ abc! + /4b%c?. 
Var — 4V/8a*bic3 + 3/6400". 
Vatbc® + 8v/b8c? — 5W/a®bec?. 


: V5 + Vash? — ~/27b5. 


. V{a + 6)? — Va? + ab — Vad? + 5B. 


Va+\/. + Ver Fa, 


V2 be ve 

Yy wv 

3a%b + 4a2c + 5aid. 
Qatyt — Saity? + Qaby}. 


165 


. , be | 
SoLuTion: Taking out the common monomial factor x7y?, we have 


2x*y? - Baty? + Qaxty? = (2 — 5xy? + Qy)a*y?. 
18. 
19. 


3aby? — 2aby3. 
2'a2b + 8a2b? — 18%a2b*. 


20. + 22 + 38. 


166 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. 
21. 3 + 12%a — 274d. 
22. /a%b? + 3»/4a%b! + 5av/9ab? — 4abv/ab?. 
23, 3/8 — 4/72 + 6V/48 — V108. 
24. VW(at+y) — Vat + xy — V8ry> + By". 
25. vF + Ways + A/a" 


102. Radicals reduced to the same order. Two radicals 
are of the same order if they have the same index. 

Radicals of different orders may be reduced to the same 
order.. This is most easily done by the use of fractional ex- 
ponents reduced to a common denominator. 

Example 1. Reduce Va and Wa to the same order. 

SoLuTion: Ja =a =a = Vai. 

AVG = Gt =e A/G. 
Example 2. Reduce 2\V/xy and 5W/xy?z? to the same order. 
SoLuTION: Q/xy = 2x*y? = Qx*y*? = 2V/x3y3. 
| Saye = 5xdyizi = 5aby'z? = 5/x%yizt, 


EXERCISES 


Reduce to the same order: 


Ley 2,44, 6. 2/m, Vxy?. 

PREM AE W od 7. 2/2, 3V 3) 
3. /xy, 20/mn. 8. wW/ay2z, 2W/axypr2?. 
4.4/5, V7. 9. 1V/ab3, 4v/abee. 
5. V/ab, W/x?y. 10. 14\/am?, 9V¥/a3m. 


103. Multiplication of radicals. We may well illustrate 
the process of finding the product of two radicals by examples. 


Che 103] EXERCISES 167 
Example 1. Multiply V3 by V5. 
Sotution: V3-V5 =3?- 5? = 15? (Law IV, Art. 89.) 
= 15. 
Example 2. Multiply 2¥/ab2m by 3~/mn. 
Sotution:  2v/ab?m - 3W/mn = 2(ab2m)* - 3(mn)? 
= 6(ab?m2n)* (Law IV, Art. 89.) 
= 6Vab?mn. 
Radicals of different orders may be reduced to the same 
order by the methods of the last article. 
Example 3. Multiply 2\/ab by 5W/a®b. 
SotuTion: 2Vab = 2V/a%b', 
5a = 5Va'b?, 
2W/ aid’ - 5V/ a4? = 10W/aib® = 10av/ab®. 


The principle used in finding the product of radicals of the 
same order may be stated as follows: 

The product of two or more radicals of the same order is equal 
to the product of their coefficients times a radical of the same 
order whose radicand is the product of the given radicands. 


EXERCISES 


Perform indicated multiplications and simplify as far as 
possible : 


1. 3\/m - 5V/an. 6. 10 -w/2. 

2. A/abe av/be. 7. 2a - 5a Wate, 
3. 2/5 -v/10 - 40/35. 8. V/4 V8. 

4, Sx -wWy?. 9. 6 -V/36 -W/5. 

5. av/b -V/ab?. 10. 5 Ve 


168 EXPONENTS, RADICALS, AND ROOTS [Caap. XI. 


The multiplication of radicals is often much more easily 
performed by the use of fractional exponents. That method 
should be used in the next ten exercises. © 


sb 


a/m -/mn + WY min?. 


av evenee ‘aoe Ls 20 eae 
SoLution : W/m +» Wmn +» oW min? =m'min'dmin? 


2 


* § 2 
=5min® = 5m + min 


oo So 
=5mvV/ mn", 


5 
6 


9 190 
= 5m + m™*ni? 


12. —/xy - 2/2". 17. V/x4y324 - Vx yz. 
13. Va: VJd- Wa’b>. 18. 3a%c} - 5a%be. 


14.-N/ 3165/3: 
15. 375 - W/2. 


19. —S8a‘b - —5a*%be?. 


20. 2a%b?ch - 9akc?. 


16. ~/128 -~/500. 21. —risi - 3r2sp. 


22. 


SOLUTION : 


23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 


(Va + 2v/b)(3Va — 5v/b). 
Va +2Vb 
3Va —5Vb 
3a + 6Vab 
— 5/ab — 10b 
3a + Vab — 10b. 
(V3 -V5)(V3 +V5). 
(Vm +/n)?. 
2/3(V3 +5 — 4/6). 
2av/b(abv/b — av/ab + 5vV/a). 
(5 — 2v/5)(3 —V/5). 
(V7 +VID(V3 -V5). 
(Bey 8 — 3/5) 
(atb? — c*)?. 
(3a? + 2b*)?, 
(at + B)3, 





Arts. 103, 104] DIVISION OF RADICALS 169 
Booey fe 8/28 —-/63)* 
34. (a? + b*)(a? — 5). 
35. (5 + 4V6 — 2v/3)?. 
36. Find the value of xz? — 47 + 1 if 2 = 2 4+. 


37; Find the value of 32? + 4x — 2 if x = Se 
= b2 — dac 
38. Find the value of az? + br + cif x = ae 


104. Division of radicals. When the dividend and divisor 
are monomials of the same order the division may be performed 


by the use of the principle 
Va 4 
Vb Vb 





66 6 = 
Example 1. oes 34/9 = 3/2. 
: aVaN NG } 
Example 2. es a ae Vi eel. /3I. 
V12 12 pees 


For purposes of computation it is usually desirable that the 
denominator of the quotient be made rational. 


A A AA) 

















Example 3. ee A 
Mx, vi V7 V7 7 
14 1-6 RA 
Example 4. ee VO ee ae 
oes Yo VV V6 6 
Example 5. 8V5+ V2_ BV5+V2) 2V543V2) _ 


De eye (2\/5 - 32) (2/5 + 30 


30 + 2/10 + 9/10 +6 _ 36 +11V10, 
20 — 18 2 


170 EXPONENTS, RADICALS, AND ROOTS [Cnap. XI. 


EXERCISES 


Perform the following divisions, obtaining results with 
rational denominators ° 


Ln Dian 8.. 6s/hiae 

DEN S80 i ND 9. 6/5 + 25. 

BON AD wb. 10. Vie 

4, ~/35 + W/5. 11. Vab + Wcd. 

5. W135 + W5. 12. Wa'm + Wam’. 
6. V/ab + Va. 13. Wax + Vy. 

7. VWabe? + Vac’. 14. 6\/150 + 5/45. 


15. (4712 — An7 Gee. 
16. (\/10 + 3/15 — 7V/35) + V/10. 


17 eo 4) Te 7 20. (/a+ Vb) + Vab. 
18.070 2415: 21. (/a +476) Se 
19. ; + ab. 22. (3 +./6) + (26 -1). 


23, 1 (28 — dao). 
24. (5\/7 — 8/11) 4/7 ey 
25. 7/2 + (2/5 — 9/2). 





ies og, Vt tvy. 
3 +V6 Va -vVy 
27. = 29. ov Oey em 
I YEeEN \/5 Seu 
PROBLEMS 


(Reduce the answers to the simplest radical form.) 
1. What is the diagonal of a square whose side is s? 
2. What is the diagonal of a square whose area is 280 
Square inches ? 


Arts. 104] PROBLEMS 171 


3. Express the volume V of a cube in terms of d, the diagonal 
of one face. Also express d in terms of V. 

4. Find the altitude and the area of an equilateral triangle 
whose side is s. 

5. Find the side s of an equilateral triangle whose area is A. 


6. Find the side of an equilateral triangle 
whose area is 420 square inches. 

7. If S is the surface and r the radius of s 8 
a sphere, then S = 47r?. Find r if S = 2000 
square inches. ; 


8. A regular hexagon can be divided into 6 ante, S 
equilateral triangles. Find the area of a regu- 


lar hexagon whose side is 10 inches. (Fig. 45.) 

9. The volume V of a pyramid of base area 
b and altitude A is given by the formula V = 
is A pyramid with a volume of 300 cubic 


feet and an altitude of 9 feet, has a square base. 
Find the length of the side of the base. BES 
10. Find the altitude and the volume 


of a regular pyramid with a square base, 
if the side of the base is s, and the lateral . 
edge 3s. 
In a regular pyramid, the perpendicular from C 
the vertex to the base meets the base at its 4. - y 


center. In Fig. 46, OA is a lateral edge. 

11. Find the altitude and volume in 
the last problem if s = 10. O 

12. Show that the total surface of the 
pyramid in Problem 10 is s*(1 + 2/2). 

13. From the formula of Problem 12 find A B 
the total surface if s = 10. 1D 

14. The pyramid in Fig. 47 is a regular c 
tetrahedron. If each edge is 10, find AD, Fie. 47 
AP, and lastly the altitude of the pyramid OP. 


Fic. 46 


172 EXPONENTS, RADICALS, AND ROOTS ([Cuaap. XI. 


The faces of a regular tetrahedron are equilateral triangles. In Fig. 47, 
AP =2 AD, where AD is an altitude of the triangle ABC. 


15. Find the volume of a regular tetrahedron whose edge 
is 10. | 


105. Square roots of polynomials. The following solution 
will illustrate the process of square root. 


Find the square root of 9a + 2a° + 22x? + 12% + 9. 


SOLUTION : 
Oat + 120% + 2247 +122 +9 | 32? 4+ 2043 
9x4 
6x? + 2x 122° + 222%? 
| 1223 + 42? 


6a? + 4a +3 Pe 
18x? + 12% +9 

EXPLANATION. (1.) The first term of the root is 9x! = 322. 
(2) The trial divisor is 2(3x?) = 62. 
(3) The second term of the root is 1223 + 6x? = 2z. 
(4) The complete divisor is 6x? + 22. 
(5) The second trial divisor is 2(8x? + 2x) = 6x? + 4a. 
(6) The third term of the root is 182? + 6x? = 3. 
(7) The second complete divisor is 6x? + 4x + 3. 


EXERCISES 


Find the square roots of the following : 


1. 25x? + 60xy + 36y’. 

2.. 4a* — 20a7b + 2507. 

3. at — 2a® + 3a? — 2a + 1. 

4. 4x* + 20x? + 292? + 102 + 1. 

5. a? + 4b? + 9c? — 4ab + bac — 12be. 

5 wy be ee 
9 5 Bite ic 15 


7. 28 — 27> Seay 22° + 5x? + 12a ee 
8. al? — 4a9 + 4a6 — a® + 2a? + 3. 


Arts. 106, 107] SQUARE ROOT 173 


106. Square roots of numbers expressed in Arabic sym- 
bols. The following solution will illustrate the process of 
extracting the square root of a number expressed in the Arabic 
notation. 


Find the square root of 419904. 


SOLUTION : 41’99’04 [648 
360000 
Trial divisor = 2 x 600 = 1200 59904 
Complete divisor = 1200 + 40 = 1240 | 49600 





Trial divisor = 2 x 640 = 1280 Mone 
Complete divisor = 1280 + 8 = 1288 10304 


EXERCISES 


Find the square roots of the following. In each of the Ex- 
ercises 5—12, find the root correct to two decimal places. 


1. 2116. 5. 85. 9. 3.1416. 
25 

2. 6084. 6. 3. 10. 2. 
5 

3. 524176. 7. 1. st 

4. 822649, 8. 016. 12. . 


107. Square roots of radical expressions of the form 
a+4/b, The square of /xtVyisa+y+ 2V/xy. Hence if 
a+/6 can be put into the form of x +y + 2vV/zy its square 
root can be found. 


Example. Find the square root of 7 + V/48. 


SOLUTION : mee ae = 7 2/12 = 443 42/4" 3S. 





Hence, te 4g a 2 + 1/3. 


174. EXPONENTS, RADICALS, AND ROOTS [Cuap. XI. 


EXERCISES 
Find the square root of each of the following : 
hen are Oe 4. 10 ++/96. 
OSG n/p: 6. 151 + 24/7. 
3. 11 + 6/2. 6. 27 =0ye 


* 108. Cube roots of polynomials. In a first course in 
algebra the process of finding the square root is explained by 
the aid of the formula 

(a + 6)? = a? + 2ab + O?. 
Similarly the method of finding the cube root is obtained by 
studying the formula 
(a + b)? = a? + 3a*b + 38ab? + B?. (1) 
Example 1. Find the cube root of 8x* + 362?y + 54ary? + 27y'. 


SoLtuTion: The work may be put in the following form. 
8x5 + 3627y + 54xy? + 27y? | Qe + dy. 


823 
Trial divisor = 3(2x)2 = 122? 36x2y + 54ay? + 27y8 
‘ Complete divisor = 12x? + 3 (22) (8y) + (8y)? 
= 127? + 18zy + 9y? B6x22y + 54ary? + 27y3 


EXPLANATION. In (1) the first term of the cube root is Ya? =a. Simi- 
larly in the present example, the first term of the cube root is 78x = 2z. 

In (1) the second term of the cube root, b, may be obtained by dividing 
3a’b by 3 times the square of the root already found, which is 3a?, and which 
is called the trial divisor. In the present example, the trial divisor is 
3 (2x)? = 122%, From 3627y + 122? = 3y, we find the second term of the 
root. 

The remainder in (1), after a3 is subtracted, may be written as the product 
of two factors, or 38a7b + 3ab? + b? = b(3a? + 3ab + b?). In the present 
example we need to find an expression of the form 3a? + 3ab + 6? where 
‘ a = 22, and 6b = 3y. This gives 

3 (2x)? + 3 (22x) (8y) + (By)? = 122? + 182y + Oy’, 
which is the complete divisor. Then 
3y (122? + 18%y + Oy?) = -- = 36r?y + 54zy? + Q7y\. 


* Articles 108 and 109 may well be regarded as supplementary work. 





Art. 108] CUBE ROOT 175 


When this product is subtracted the remainder is zero. The process is 
then complete and the cube root is 2% + 3y. 

It should be noticed that the complete divisor is formed by adding to 
the trial divisor an expression of the form 3ab + b?. 


Example 2. Jind the cube root of 
8x° — 36275 + 6621 — 6323 + 332? — Ox +1. 


SOLUTION: 82° — 36z° + 6624 — 632° + 332? —- 9x7 41 | 27? —- 327 +1 
a ad 
1st trial divisor, 
3 (2x7)? = 1224 | — 3625 + 6624 — 6323 

1st complete divisor, | 

1224 — 1843 + 92? | — 3625 + 54a4 — 2723 
2nd trial divisor, 
3 (2a? — 3x)? = 1224 — 3623 + 272? | 1204 — 3623 + 332? — 97 +1 
2nd complete divisor, 
1224 — 362° + 332? —- 9x +1 1224 — 362° + 332? — 9x +1 





The first two terms of the root are found as in Example 1. 
To find the third term we proceed as in finding the second term. 
Three times the square of the part of the root already found is 


3 (2x? — 3x)? = 1224 — 3623 4 272?, 


which is the trial divisor. Dividing, we find the next term in the root 
to be 1. 
To find the complete divisor we must add to the trial divisor 


3(22? — 3x)-1 + 1? = 62? — Ox +1. 
The complete divisor is then 
1224 — 362° + 332? —- 97 + 1. 


Multiplying this by the last term of the root and subtracting, the remainder 

is zero. The process is then complete and the cube root is 2a? — 3x + 1. 
If the root contains more than three terms, the succeeding terms can be 

found by continuing the process used in finding the second and third terms. 


EXERCISES 
Find the cube roots of the following expressions : 
1. 273 + 272? + 9x 4+-:(1. 
2. x? + 6a?y + 12ry? + 8y’. 
3. 8a° — 36a2b + 54ab? — 276%. 


176 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. 
p> — 3pty® + 3p2y® — . 
64a°b? + 144a7b? + 108ab + 27. 

1 — 15m?2x +. 75m4*x? — 125m 523. 

- 216x%y> + 108m2rty2 + 18m*x2y +m. 
m> + 3m? + 3m + §. 

1: — 624+ Qla? — 449*:-+ 632+ — bat ease 

10. b® — 3b + 5b? — 36 — 1. 

11. 8z!? — 36z!° + 662° — 632° + 33z4 — 9a? + 1. 

12. 1252° + 1502y + 2252? + 60zry” + 180zry + 135” + 8y? + 
36y? + 54y + 27. 

* 109. Cube roots of numbers expressed in Arabic symbols. 
The process used in finding the cube roots of polynomials can 
be applied to numbers expressed in Arabic symbols. This will 
be illustrated by an example. 

Example. Find the cube root of 185193. 


ret bacspe wil ahh lho 


SOLUTION : 185/193 |57 t = 50, 
125000 = @ u=7. 
Trial divisor = 3t? = 3 - 50? = 7500 60193 contains 3t?u + 3tu? + wu. 
situ =3-50-7 = 1050 
u? = 7% = 49 
Complete divisor = 
3? + 3tu+u? = 8599 60193 = 3f?u + 3tu? + u3. 


EXPLANATION. (1) The given number is divided into periods of 3 figures 
each to determine the number of orders in the cube root. 

(2) Since the number contains 2 such periods, the root contains two 
orders, tens and units, which may be represented by ¢ and u, respectively. 
Hence the cube is in the form | 

(¢ + u)? = #8 + 3éu + 3tu? + Ww, 
(3) The largest cube of tens contained in the number is 
# = 50? = 125000. 


(4) After @ is subtracted the remainder contains 3f2u + 3tu? + u3. 
* Finding cube roots of numbers may well be delayed until the next 


chapter where it is done much more simply and easily by the aid of 
logarithms. 


Art. 109] = EXERCISES AND PROBLEMS 177 


(5) The trial divisor is 3 times the square of the root already found, 
which is 3¢? = 3 - 50? = 7500. 

(6) The next figure in the root is found by division to be 7. 

(7) The complete divisor is 


3t? + 38tu + uw? =3 - 50? +3 .- 50-7 +7? = 8599. 
(8) Multiplying the complete divisor by the new figure in the root we 
have 7 x 8599 = 60193. Subtracting, the remainder is zero. Hence the 
process is complete, and the cube root is 57. 


(9) If the cube root contains more than two orders, the remaining orders 
may be found by repeating steps 5, 6, 7, and 8. 


EXERCISES 
Find the cube roots of the following : 
1. 13824. 5. 636056. 
2. 110592. 6. 21952. 
8. 421875. (fm F334 0559 be 
4. 804357. 8. 4913. 


MISCELLANEOUS EXERCISES AND PROBLEMS 


In Exercises 1-6 rationalize the denominator and find the 
values of the fractions correct to two decimal places. 











x Be. ‘ 2/5 —V7 
1472 V5 +377 

; ae Rae 
VJ/3 +/5 37 

3. 5+V3 | 6. aS FA Od eae 
Oh 24/3 5/2 — 4/3 


7. Find the value of x? — 62 + 2 when 2 = 3 + 2v2. 
8. Find the value of x? — 4a — 1 when x = 2 -vV/5. 
V19 
oe 
10. The area of a square is 14 + 6\/5. Find the side correct 
to two decimal places. 


9. Find the value of 3x? — 27 — 6 when z = 1 + 


178 EXPONENTS, RADICALS, AND ROOTS [Cuap. XI. 


11. Find the side of a square whose area is 998001 square 
feet. 
12. Find correct to two decimal places the radius of a circle 
whose area is 50 square inches. 
13. In Fig. 48, T is a right triangle, and 71, T:, and 73 are 
equilateral. Prove that 7; = T, + 73. 
14. The circumference of a circle equals 
the perimeter of a square which is 4s. 
Find the difference in their areas in terms 


of s. 
e 15. Assume a given unit of length and 


Fic. 48 construct geometrically /5. 


‘SuccEstion. Use codrdinate paper. Construct a right triangle with 
legs 1 and 2. Then the hypotenuse equals V1 +4 =-+V/5. By actual 


measurement the value of 5 can be determined correct to tenths, or even 
hundredths, if the scale is taken sufficiently large. 


16. Construct the following square roots: 7/2, 10, 17, 
x/3, V 20. 

Suacestion. In constructing 3 construct a triangle with base 1 
and altitude 2. 


17. Plot carefully the graph of y? = x for values of y from 1 
to 10. Solve the equation for y. This gives y =z. If now 
for given values of x we read off from the curve the values of 
y, we thus obtain the square roots of those values of z. By the 
aid of this curve find the approximate value of the square root 
of 2, 3, 10, 16, 38, and.75. 

18. Similarly from the curve y = 2° plotted from z=1 to 
x = 5 obtain the cube root of 2, 3, 27, 40, 65. 


CHAPTER, XII 
LOGARITHMS 


110. Definition of a logarithm. In the expression 103 = 
1000, the exponent 3 is called the logarithm of 1000 to the 
base 10, and is written 3 = logi1000. 

In general, if a* = y, (a>0, a #1), then 2 is said to be the 
logarithm of y to the base a, which is written x = logyy. 

The two equations 

. iy; 


and t= l0g,y 


are thus equivalent. That is, the logarithm is an ex- 
ponent. 

Without using symbols, the logarithm of a given number 
may be defined as the exponent of the power to which a 
number called the base must be raised to equal the given 
number. 

In what follows, we shall assume that the laws of exponents 
(Arts. 89-94), which apply to rational exponents, are valid for 
any set of exponents with which we are concerned. 


111. Numbers expressed as powers of given numbers. 
We know how to express many numbers as powers of a given 
number. These same relations may be expressed by means 
of logarithms. | 


Thus, 16 = 42, and 2 = log,16; 8 = 4°, and : = log, 8 ; 4 = 4% and 


1 1 = 1 
—3 = logiea ; 100 = 102, and 2 = log100 ; ioe 10“1, and —1 = logw7o: 


180 LOGARITHMS [Cuap. XII. 


EXERCISES 
1. logs36 = ? logiwl00 = ? logs = ? logel28 = ? 
logs81 = ? 
2. Fill out the following table : 


Base Number | Logarithm 
10 10 | 
a 81 4 
3 8 
8 : 
10 —3 
10 1 


—E7 





3. Write the following in the notation of logarithms: (a) 
452 =.10'5!- = (6) 5.1 = 10°: (6) 6470 
DADO Tey SOT 

4. Write the following using exponents: (a) logi2 = .38010 ; 
(b) logi99 = 1.9956; (c) logw:34.2 = 1.53840; (d) logis7000 = 
3.8451. 


112. Characteristic and mantissa. The integral part of a 
logarithm is called the characteristic and the decimal part the 
mantissa. 


Thus, in log:9245 = 2.38892, the characteristic is 2 and the mantissa is 
3892. 


The characteristic may be negative. 
Thus, logi.01 = —2. 


It is usually convenient to keep the mantissa positive even 
though the logarithm is negative. 
Thus, logio+ = —.6990. But this may be written 1.3010, which has a 


positive mantissa, and a characteristic —1. The minus sign is placed 
over the 1 to indicate that the characteristic alone is negative. 


Arts. 113, 114, 115] PROPERTIES OF LOGARITHMS 181 


113. Logarithm of a product. The logarithm of a product 
equals the sum of the logarithms of its factors. 


. Proor: Let logau = x and logav = y, 

then a® =u,a¥ =v, (Definition of logarithm) 
and uv = arty, (Art. 89) 
Hence, loggw =x+y, 
that is, log,uv = logau + logy. 


Similarly, loga(ww) = logvu + log + logaw, 
and so on for any number of factors. 


Example. logio(79 x 642) = logiw79 + logi.642. 


114. Logarithm of a quotient. The logarithm of a quotient 
is equal to the logarithm of the dividend minus the logarithm of 
the divisor. 


Proor: As above, let logaw = x and logy = y. 


Then a* =-u, and a¥ = 2, 
and Sergent 
v 
Hence, log =“—Y, 
; U 
that is, loge, = logau — logav. 


Example. loging; = logw245 — logy912. 


115. Logarithm of a power. The logarithm of w’ is equal 
to v times the logarithm of w. 


PROGR eLCa et — 102,0, OF a*.= u. 
Then Pe (tt) a"? (Art. 89) 
Hence, logaw’ = vx = v logyu. 


Example. logic (455)? = 3 log0455. 


182 LOGARITHMS [Cuap. XII. 


Making v = n, and then v = we have respectively, 


(a) The logarithm of the nth power of a number is n times the 
logarithm of the number. 


(b) The logarithm of the real positive nth root of a number is 
the logarithm of the number divided by n. 


EXERCISES 


Express the logarithms* of the following expressions in 
terms of the logarithms of integers: 














LV 2 
oa lt 
ar cipky 
SOLUTION : 
ete log 17 + log W/12 — log 7? — log 11? (Arts. 113, 114, 115) 
= log 17 +4log12 — 2log7 — tlog11 (Art. 115). 
48 3V5 
2. log 05 6. log WF 
2/180 54. ¥/1.9 
Sal Teel . 
19 °° 6 te 
12? - 325 945 
4. lo ae =e 
ea 6 8. logy sii 
(58)8 a 
5. | : Suit 
11/2 el: (V3 : 


(vs ()) 


* When in a problem the same base is used throughout, it is customary 
not to write the base. 


Arts. 115, 116] COMMON LOGA 


Express the logarithms of the foll 
of the logarithms of prime factors : 


3 





11. lo =" 
2 cVd 
SoLuTIon : log od, loga + 8logb — 
cVd 
12. log le Bt 
fe 
vn 16 
“iG i a 
: n(m — 2) 
its 





14. log Va 


18. 


RITHMS 183 


owing expressions in terms 


loge — $logd. 


log(a? — b?). 

° log od oe 
Vc(ax + b) 

ta? 


vy 
log(a? + 6). 


log 


Express each of the following sums and differences as the 
logarithm of a single expression: 


19. log m —1 log n + 2 log z. 


. 1 “'/ mx? 
Soxution : log m —tlogn +2logz =logm —logv n + log x? = log Wa 
n 


_ 20. 
ri i 
22. 
23. 
24. 
25. 


5 log 3 + 2 log 7 — 4 log 95. 

log 48 + 3 log a — 2 log b — 3 log 15. 
a log x — b log y. 

log(a — 6) + log(a + 6). 

log 1 — log 15. 

2 log(a? + b?) — 2 loga — 2 log b. 

116. Common logarithms. As 10 is the base of our number 
system, it is convenient to take 10 as the base of the logarithms 
used in ordinary computation. These logarithms are known 
as the common logarithms, or the Briggs system of logarithms. 


In the following discussion of common logarithms, log x is 
written as an abbreviation of logioz. 


184 LOGARITHMS [Cuap. XII. 


117. Determination of the characteristic of a common 


logarithm. 
Since 10? = 1000, log 1000 = 3; 


” 107 = 100, log JOU3eree 


* 10' = ©10;ior \ oe 
BAG Oot te 1, log 1303 
7 107 =. 1 log 2) ae 
” 10°? = (Ol, log. 0 ae 


? 107° = 001, loe-.00I== re 
and so on. 

So far as these powers of 10 are concerned it may be observed 
that when a number increases, its logarithm increases. As- 
suming this to be true for all positive numbers we have, for 
example, 


log 100< log 400< log 1000, 


or : 2< log 400<3. 
Hence the characteristic of log 400 is 2. 
Similarly, log.01 <log.021 <log.1, 

or —2<log.021< —1. 


Hence the characteristic of log.021 is —2, as the mantissa 
is always taken positive. 

Exercise. Find the characteristic of the logarithm of each of the 
following : 15, 348, 8000, 965, .3, .045, .00961, .002. 

The above illustrations show that if a number is between 
1 and 10 the characteristic of its logarithm is 0; if between 
10 and 100, the characteristic is 1; if between 100 and 1000 
the characteristic is 2; and soon. Hence, we have the eee 


- rule: 


The characteristic of the logarithm of a number great than 
1 is one less than the number of as in the integral part of the | 
number. 


Thus, the characteristic of log 34.5 is 1; of log 8964.701 is 3; and so on. 


Arts. 117, 118, 119] MANTISSA OF A LOGARITHM 185 


We see also that if a number is between 1 and .1, its charac- 
teristic is —1; if between .1 and .01, its characteristic is —2 ; 
if between .01 and .001, its characteristic is —3; and so on. 
Hence, we have the general rule: 

The characteristic of the logarithm of a number between 1 and 
0 is negative and numerically one greater than the number of zeros 
between the decimal point and the first significant figure. 

Thus, the characteristic of log.0345 is —2; of log.000479 is -4; and 
so on. 

118. Mantissa of a common logarithm. From the table, 
p. 187, we shall see later how to find that 

log 5682 = 3.7545. 
Then by Art. 114, log 568.2 = log — = log 5682 — log 10 
= 3.1045 — ] = 2.7545. 
Similarly, log 56.82 = 2.7545 — 1 = 1.7545, 
log 5.682 = 1.7545 — 1 = .7545, 
log 5682 = .7545 — 1 = 1.7545, 
log .05682 = 1.7545 — 1 = 2.7545. 


I 


Hence, the mantissa of the common logarithm of a number is 
independent of the position of the decimal point in the number. 
In other words the common logarithms of numbers which 
contain the same sequence of figures differ only in their charac- 
teristics. Hence tables of common logarithms contain only the 
mantissas, and the computer must find the characteristics by 
the rules of Art. 117. 


For example, the table, p. 187, gives .8722 as the mantissa for the se- 
quence of figures 745. We know then immediately that log 745 = 2.8722, 
log 7.45 = .8722 and log 00745 = 3.8722. 


119. Graph of y = logy (a>1). <A general notion of the 
values of logarithms of numbers can be easily fixed by refer- 
ence to the graph of y = log,x. This graph is also the graph of 





9978 


188 LOGARITHMS [Cuap. XII 


x =a’. In making the graph (Fig. 49) we took a = 10, but the 
general form of the graph would not be changed if a were given 
any other positive value greater than one. In particular, the 
curve crosses the z-axis at a point a distance 1 from the origin, 
for any value of a. If the student retains this picture, he 
should find it easy to keep in mind the following facts when 
the base is greater than unity. 

(1) A negative number does not have a real number for its 
logarithm. 


Y 





Fig. 49 


(2) The logarithm of a positive number is posi- 
tive or negative according as the number is greater 
or less than 1. 
(3) If x approaches zero, log x decreases without limit. 
(4) If x increases indefinitely, log x increases without limit. 


120. To find from the table the logarithm of a given number. 


EXAMPLES 


1. Find the logarithm of 821. 


Glance down the column headed N for the first two significant figures, 
then at the top of the table for the third figure. In the row with 82 and 
the column with 1 is found 9143. 


Hence, log 821 = 2.9143. 
2. Find the logarithm of 68.42. 


This number has more than three significant figures, so that its logarithm 
is not recorded in the table. It may, however, be obtained approximately 


Arts. 120, 121] USE OF TABLES OF LOGARITHMS 189 


from logarithms recorded in the table by a process called interpolation. In 
our use of this process, it is assumed that to a small change in the number, 
there corresponds a change in the logarithm which is proportional to the 
change in the number. This assumption is called the principle of propor- 
tional parts. Asin Example 1, we find that the mantissas of 6840 and 6850 
are 8351 and 8357, respectively. The difference between these two man- 
tissas is 6. Since 8642 is two tenths of the interval from 6840 to 6850, by 
the principle of proportional parts, we add to 8351, 

u 0.2 x6 =14+4. 

Hence, log 68.42 = 1.8352. 

3. Find the logarithm of .0478. 


From the table the mantissa is 6794. Hence the logarithm is 2.6794. 
For purposes of computation it is convenient to write this logarithm 
in the form 8.6794 — 10. This form will be used hereafter. 


121. To find from the table the number that corresponds 
to a given logarithm. 


EXAMPLES 


1. Find the number whose logarithm is 2.4675. The mantissa 4675 
is not recorded in the table, but it lies between the two adjacent mantissas 
4669 and 4683 of the table. The mantissa 4669 corresponds to the number 
293 and 4683 corresponds to 294. The number 4675 is 3% of the interval 
from 4669 to 4683. By the principle of proportional parts, the sequence 


of figures whose mantissa is 4675 is 2930 + 1 x 10 = 2934+. 
Hence, 2.4675 = log 293.4. 
2. Find the number whose logarithm is 9.3025 — 10. 
From the table, log 0.2000 = 9.38010 — 10 
log 0.2010 = 9.8032 — 10 
Difference = 0.0022 
(9.3025 — 10) — (9.3010 — 10) = 0.0015 
By the principle of proportional parts, the number is 


0.2000 + = x 0.0010 = 0.2007. 
It should be noted that whenever a computation is performed by means 
of four-place logarithms, the answer should not be given to more than four 
significant figures. 


190 LOGARITHMS [Cuap. XII, 


EXERCISES 
Find, from the table, the common logarithms of the following : 
UE atary: 5. .00289. | 9. 325000. 
HANEY 6. 3642. 10-201 
3.30. (fe Tyla 11. .009793. 
4. 6.25. 8. .1414. 12. 68.07 


Find, by means of the table, the numbers whose common 
logarithms are the following : 


13. 1.5502. 17. 1.3478. 21. .9762. 
14. 3.9031. 18. 8.7542 — 10. 22. 4.1572. 
15. 9.7348 — 10. 19. 3.5740. 23. 9.6990 — 10. 
16. .8882. 20. 7.2182 — 10," 2an 


122. Computation by means of logarithms. The application 
of logarithms to shorten calculations depends upon the proper- 
ties of logarithms given in Arts. 118-115. By means of loga- 
rithms laborious multiplications can be replaced by additions 
and subtractions; and involution and evolution can be replaced 
by multiplication and division. 


EXAMPLES 


1. Find the value of N = eet to four significant figures. 


log 85 = 2.9294 
log 43.67 = 1.6402 
log(85 x 43.67) = 4.5696 
log 596 = 2.7752 
log N = 1.7944 
N = 62.29 
In using logarithms, much time is saved and liability of error is de- 


creased by making a so-called form for all the work before using the table 
at all. 


Art. 122] 


Thus, in Example 1, the ‘‘form” is 
log 85 = 


2. Make a form for evaluating N = 


log 43.67 


log (85) (43.67) = 


log 596 


log N = 
N= 


(478.9)! - 


V/834 


log 478.9 = 


log (478.9) 
log.0596 


logw/.0596 
log (478.9) #~/.0596 





log 834 


logW834 
log N 


EXERCISES AND PROBLEMS 


APPLICATIONS OF LOGARITHMS 


0596. 


Evaluate to four significant figures by logarithms: 


ii 
2. 


3. 


4. 


139 x 971. 


49.1 x 12.37. 


4000 


982.7. 


1.5692. 
V2. 
10. 


1.06 X 73.9, 


19.14 
4/0537. 
v.02. 


10. 


11. 
12. 
13. 


14. 


V3 


Ypres 
~.0965 


538,947 x »/.0003. 


( NS 
35.14 


Wes x 3Be 
86.75 


. o44y/ SE 


Z2tcLe 


73. 140m 


191 


192 LOGARITHMS [Cuap. XII. 


: . 2 
16. The area of a circle of radius 7 1s mr?.. Use = for 7. 


(a) Find the area of a circle whose radius is 17.3 inches. 
(b) Find the radius of a circle whose area is 456.7 square 
inches. 


17. The volume of a sphere is mrs 


(a) Find the volume when r = 12.7 inches. 

(b) Find r when the volume equals 148 cubic inches. 

18. Find the weight in tons of a cylindrical marble column 
whose length / is 12.7 feet, and whose radius r is 2.9 feet, if 
marble is 2.7 times as heavy as water, and a cubic foot of 
water weighs 62.5 pounds. The formula for the volume of a 
cylinder is V = wr’. 

19. How many acres are there in a triangular field whose 
sides are 30 rods, 14.7 rods, and 23.5 rods? 


If a, b, and c are the sides of a triangle, its area is Vs(s — a) s — b)(s — 6), 


a+b+ec 
where s =) a ae 


20. The amount of P dollars at rate r for n years com- 
pounded annually is A = P(1 +1)". What is the amount of 
$50 for 10 years at 4%, compounded annually ? 

21. The side S, of a regular pentagon inscribed in a circle of 


radius r is S =5y/ 10 _ 9./5. Bind S if peat 


Co 








Fig. 50 Fic. 51 


22. The side of a regular pentagon is 10 inches. Find the 
radius of the circumscribed circle. 


Arr. 122] PROBLEMS 193 


The portion of a sphere cut off by one plane is a special kind of spherical 
segment. (See Art. 45, Fig. 17.) The section made by the cutting plane 
is called the base of the segment. In the figure the circle ACB is the base; 
MO’ is perpendicular to this circle; and MO’ is the altitude of the segment. 
The volume V, of a spherical segment of altitude h, in a sphere of radius 


a8 .V. = hel r -5) 


23. Find the volume of a spherical segment of altitude 2.67 
inches in a sphere of radius 6.09 inches. 

24. Show that the foregoing formula for the volume of a 
spherical segment can be deduced from the formula given in 
18, Art. 45. (Note that here b = 0.) 

25. An auger hole 1 inch in diameter is bored through the 
center of a croquet ball 4 inches in diameter. What is the vol- 
ume of the wood removed ? 

26. Estimate the weight of a cork ball 6 feet in diameter and 
then calculate its weiglit in pounds. A cork whose volume is 1 
cubic inch weighs 0.139 ounces. 

27. First estimate and then calculate the increase in weight 
of the ball in Problem 26 if the radius is increased 1 inch. 

28. The Dutch paid $24 to the Indians in 1614 for Manhattan 
Island. What would this $24 amount to in 1914 if it had been 
on interest at 5% compounded annually ? 

29. What would be the amount to-day of 1 cent which 1900 
years ago was placed on interest at 6%, compounded annually ? 
Find the radius in miles of a sphere of gold which has this value. 


A cubic foot of gold weighs 1206 pounds avoirdupois; 1 such pound 
contains 7000 grains; and 23.22 grains of gold are worth a dollar. 


Historical note on logarithms. Logarithms were invented by Baron 
John Napier (1550-1617) of Scotland. 

The base used by Napier was not 10; but, soon after the publication of 
the invention, Briggs, a professor in London, visited Napier, and it seems 
probable that they both saw the usefulness of a table of logarithms to the 
base 10. Briggs devoted himself to the task of making such tables. For 
this reason, logarithms to the base 10 are often called Briggs’ logarithms. 


194 LOGARITHMS [Cuap. XII. 


An early incident of the visit mentioned above is frequently quoted. It 
appears that Briggs and Napier were very anxious to have a visit, but that 
after meeting, about a quarter hour was spent, each beholding the other 
without saying a word. Finally Briggs spoke as follows: “My lord, I 
have undertaken this long journey purposely to see your person, and 
to know by what engine of wit or ingenuity you came first to think of 
this excellent help in astronomy, viz., the logarithms; but, my lord, 
being by you found out, I wonder why nobody found it out before, when 
now known it is so easy.” 


Art. 122] REVIEW EXERCISES AND PROBLEMS 195 


REVIEW EXERCISES AND PROBLEMS ON CHAPTERS IX-XII 
1. Solve the following quadratic equations as indicated: 
(a) x? + 6x — 27 = 0, by factoring. 
(b) y? + 10y — 3 =0, by completing the square. 


(c) 2? — 2z = 11, by formula. (WISCONSIN) 
2. Solve (x —y)? + (@+y)? = 10, 
(xz —y)(@ +y) =3. 
was: ga tag ES 
3. Simplify a’. ght |™ 


4. Determine for what values of a the roots of 
4y? —2(a —3)x +1 =0 


are, (1) real and equal, (2) real and unequal, (3) imaginary. 
(Mass. INSTITUTE) 


5. Reduce to a single number by means of logarithms 














000743 x 23.5625 + ~/4000. (YALE) 
Toe | 
6. Compute ert ae by means of logarithms. 
(PENNSYLVANIA) 
ge 2 on 3 Die, 
7. Solve ee AY Gs 
(ILLINOIs) 
8. Solve to three significant figures 
x — 1.62 — 0.23 = 0. (HARVARD) 
aye Sie ie ee a 
9. Simplify (ay) (ay) and express the result without 
using negative exponents. (CASE) 
10. Solve the simultaneous equations by two methods: 
a*+ay+y’? = 39, 
er+y = —2. (WISCONSIN) 


11. Simplify the following 
(a) 2/3 — V/12 + V9. 
(b) W2x V4 x V3. 


1 \2n 
(c) \ a. bin ; (SHEFFIELD) 


196 LOGARITHMS [Cuap. XII. 


12.55 3 3 
Sie and n = (.0125)* (423)’, 


given log 125 = 2.0967, log 126 = 2.1004, log 422 = 2.6253, log 423 = 2.6263. 


12. Calculate log m and log n, when m = 











(NEBRASKA) 
: og 3 3 
13. Find ey a log - 
14. How many digits in the number 2!” ? 
2 a(l+22)  b (8% —1)\ _ 
15. Solve Coe (5 (1 2Sz).2 8 oe 3] = 0. (HARVARD) 


16. (a) Find without solving, the sum and product of the roots of 
—32? +127 +11 = 0. 


- (b) Find an equation whose roots are : and 3. 


3 
(c) If 2 +3 -—1 is one root of a quadratic equation, what is the other 
root? — : (Missouri) 
I 
17. Solve x+ % =l,y+ ma 4, (PRINCETON) 


18. (a) For what value of c does the equation 67? + 8% +c = 0 have 
equal roots? 

(b) Solve 162? = 7a — 1. (ILLINOIS) 

19. Express without radicals in simplest form 


(a) Rie as (b) (//azat) (NEBRASKA) 


20. Find a logarithmic expression for the amount of $1.00 placed for 


10 years at 5% compound interest compounded semiannually. 
(SHEFFIELD) 
21. For what value of b will the simultaneous equations 


16x? + 25y? = 400, 


4 
y= Fed + 6. 
be satisfied by only one pair of values of x and y. (HaRVARD) 


22. Give the numerical value of 


(a) 4°, (b) (-8)*, (c) 3-2, (d) 5°, (WISCONSIN) 
23. Solve the equation 
2 log 6 — 5log2 = 3. (CORNELL) 
24. The number of revolutions per minute of a water-wheel is given 
by the formula 
Ht 


Art. 122] REVIEW EXERCISES AND PROBLEMS 197 


where H is the fall of the water in feet, P the horse power. Calculate n for 
WH = 20, P = 50; also for H = 20, P = 100. 

25. A pays $420 for the rent of a farm. He lets all but 4 acres to B, 
charging $2.50 per acre more than he paid, and receives from B as much 
as he originally paid. How many acres in the farm? |= (SHEFFIELD) 

26. A boat’s crew, rowing at full speed, rows 2 miles down stream and 
back again in 35 minutes. Rowing at half full speed it takes 80 minutes 
to go over the same course. Find the rate of their rowing and the rate of 
the stream. 


27. Perform the indicated operations : 

(a) (ras) - (r? 73). 

(b) 12-3 + 11-8, 

(c) 10°23 + 101-5, (WISCONSIN) 
28. Find the base of a system of logarithms in which log } = 1.8000. 


29. Simplify | (aria. (a)? |. (PRINCETON) © 


30. A man bought a flock of sheep for one hundred dollars. If he had 
bought five less for the same sum, they would have cost one dollar more 
per head. How many did he buy? (CasE) 

31. A park is 120 rods long and 80 rods wide. It is decided to double the 
area of the park, still keeping it rectangular by adding strips of equal width 
to one end and one side. Find the width of the strip. (Missourt) 

32. Given the equation 

x? + bx —-18 = 0. 
If one root is — 9, what is the other, and what is the value of 6? 
(CALIFORNIA) 
33. Express the solution of 4°*-! = .57~! by using logarithms. 
(SHEFFIELD) 

34. Both the numerator and the denominator of a fraction are increased 
by their squares. The resulting fraction reduces to 4. The sum of the 
numerator and denominator is 5. What is the fraction? (ILLINOIs) 


35. Find the tenth root of one tenth. 
36. Find the value of y from the equation 
1 =t (ete 
cee 3 VS (HarvARD) 
37. The plate of a mirror is 18 by 12 inches, and it is to be framed with 


a frame of uniform width whose area is to be equal to that of the glass. 
Find the width of the frame. (CasE) 


198 LOGARITHMS [Cuap. XII. 


38. The joint stock of two partners was $2000. A’s money was in trade 
12 months and B’s 15 months. A received $1272, capital and profit. B 
received $860 capital and profit. What was the capital of each ? 

39. What is the price of eggs when 2 less for 24 cents raises the price 
2 cents a dozen? (Y ALE) 

40. Which is the larger 25 or V6? Prove your answer. 

41. When a body is thrown vertically upwards, with an initial velocity 
of v feet per second, the distance d from the point of departure after ¢ 
seconds have passed, is d = uot — Zgt?, where g = 32.2. If a body is thrown 
vertically upwards with an initial velocity of 64.4 feet per second find the 
time when d = 48.3 feet. Explain the two answers. (ILLINOIS) 

42. What would be the answer to Problem 41 if, as is the case on the 
moon, the value of g were 5.4? 


43. Given xt + y% = al, 
pe 
Pri at 
x\i/y — px 
o=4() eee 

pte 

q 
Express R in terms of x and y and in as simple a form as possible. 


(DarTMOUTH) 


44. Which yields more, a sum left 10 years at 5%, or 5 years at 10%, 
interest compounded annually? What is the difference for $1000? 

45. A rectangular field is three times as long as it is broad. The shorten- 
ing of the field by 3 rods and the broadening by 4 rods has the effect of in- | 
creasing the area by one-half the original area. Find the original area in 
square rods. - (PRINCETON) 


46. Simplify E 





(ILLINOIS) 


cold 

(RES 

Lely 
Spe 
~~ 
or 

] 

bo 

oO ee 


47. $4400 is invested in two parts and at different rates of interest, so 
as to give the same income. If the first part were invested at the second 
rate, the income from this part would be $100 ; but if the second part were — 
invested at the first rate, its income would be $144. What are the rates 
of interest ? 

48. A rectangle is 6 inches wide and 8 inches long. Keeping the width 
the same, how many inches must be added to the length to make the diag- | 
onal 1 foot long? (Correct to hundredths of inches.) (STANFORD) 


Art. 122] REVIEW EXERCISES AND PROBLEMS 199 
49. Simplify 
V(il+a)1+y) -V0d -2)0 -y) 
Va +a +9) +V0-20-y) 
50. At simple interest, $300 amounted to $405 at a certain rate and 


time. If the time had been one year longer and the rate 1 % less, it would 
have amounted to $408. What were the time and the rate? 


51. Find the value of x in the following equations : 








ae . 
(a) Gi gee a2; (b) log 5 + log(8a — 2) =3. (DARTMOUTH) 


CHAPTER XIII 
EQUATIONS INVOLVING RADICALS 


123. Solution of equations containing radicals. Certain 
equations in which the unknown is involved under the radical 
sign can be reduced to equations of the first or second degree. 
The following examples illustrate the method of solving some 
of the more simple but typical equations in which such reduction 
can be made. 


EXAMPLES 
1. Solve the equation V3z +1 = 5. 


SoLuTION : Squaring both members, 


3-8 L-= 25: 
Solving for z, doo aa 
CHECK : V/25 = 5. 


It should be recalled that 25 = +5, and does not equal +5; that is, 
when no sign precedes the radical the positive value of the root is to be 
taken. If both positive and negative roots are meant, we shall write 
both signs before the radical. 


2. Solve /4¢ +5 +2Va2 -—3 = 17. 
SoLuTION : Transposing, 
V42 +5 -17 = -2Vz —3. 


Squaring, Ay +5 — 34\/4e +5 + 289 = 4x — 12. 
Transposing and simplifying, V/4x +5 = 9. 
Squaring, 44 +5 = 81. 
Solving for 2, ieee yt 
CHECK : V/81 + 2V16 = 17, 


or, Wee aif 


Art. 123] EQUATIONS CONTAINING RADICALS 201 


8. Solve Vz —-2 —-V2x4+3 =1. 
SoLuTIoNn : Transposing x + 3 and squaring, 


e2—-2=24+3+42V¢243 +1. 
This reduces to Vx+3 =—8. 
Solving, x = 6. 
But 6 is not a solution of the given equation. In fact, the given equation 
has no solution. 
4. Solve vV/52 —4 =V2xe +1 +1. 


SOLUTION : Squaring and transposing, 32 ~~ 6 = 20/22 + I. 
Squaring and transposing again, 9x? — 447 + 32 = 0. 
Solving, a= 4, or-§. 


The value 4 satisfies the given equation, but $ does not. We notice, 
however, that § does satisfy the equation if the signs of the radicals are 
taken negative. 


5. Solve 2V/z 4+ 2=vVJ2x4+1— 2. 


SoLuTION : Squaring and transposing, —4V/ 2.41 = 37 +3: 
Squaring and transposing again, 9x? +27 —7 = 0. 
Solving, x = —l,orZ. 


Neither of these values is a root of the given equation. 


The above examples illustrate that : 


(1) Two expressions involving radicals may not be equal 
for any value of the unknown, as in Example 3. 

_(2) Extraneous roots may be introduced by squaring, as in 
Examples 4 and 5. 

(3) Results must always be checked. There is no other way 
to determine whether or not a result is a root of the given equa- 
tion. In Examples 3, 4, and 5, results are obtained which are 
not roots of the given equation. 

Steps in the solution. In solving equations containing radi- 
cals it is usually convenient to proceed as follows: 

(1) Isolate the radical; that is, place it by itself in one 
member of the equation. If more than one radical occurs, 
isolate the most complicated one. 

(2) Square both members of the equation. 


202 EQUATIONS INVOLVING RADICALS [Cuap. XIII. 


(3) Ifa radical remains, isolate it and square again. 
(4) Solve the resulting equation. 
(5) Check the results. 





EXERCISES 
1. V2 +3. = 2. 5. Va Sele 
2. /34 — 5 eget): Hint: Cube each member. 
3. Voto =U, 6. Vx —447= 9, 
AlN een ea 7. (24 — 6 )iseeone 
8: 10 Se Va: 
9 Yet 15 =Vr4+1+2. 


10. 0/90 4.6 — 8\/e 011 ee 

11. 2 -4-—V/27 —5 = 0. 

12. /10+2-V10—-2 =2. 

13. /2xr + 1-2 =}. 

14. 2/2 —3 4+ 3V 2 —2 = 13% = 80. 
16. Ve+1l=V9—2—-Vrc $e 

16. 2+5/x = —6. 

17. /3a+1=V7/5¢ —4 —- Va —7. 
18. \/2 —7 — 524 = see 
ViGgtel aee 
Va—-5° 2 
VEtVv2e 
Vx +2 
Va = 2 +9 _ 
/ 30? — 2 


oP ene , oe 
22. a Brae: Te: 





19. 


20. 1 





21. 





| 


Art. 123] EQUATIONS CONTAINING RADICALS 


23. 


24 


25. 
26. 


27. 


28. 


29. 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


MV/X — p= NvV/x — G. 
VatetvVa—-2 = vV2a. 
a+ Ja? — 2? =%2. 


V/m—-xtVr—-n=Vm—n. 


Bert Vb Ae ain Va 
V8 +V Te —8 | 





ee Se 4/7 — 6 
> eis z—n 15 
r—n re: 


a 4/ ne = 30 = 4b := ——__——. 














Solve V = ; wr for r. 
1/38? 


4 for s. 





Solve A = 





_ mgl 
Solve s = ay. for r 


Solve d = .02758VD -1- WP for P. 
Solve V = é (B+6+~/Bb) for B. 


Solve A = P(1 +7) for r. 


203 


CHAPTER XIV 
EQUATIONS IN QUADRATIC FORM 


124. Quadratic form. Any equation is in quadratic form 
with respect to an expression containing the unknown, if we 
may replace such expression by a new letter, and obtain a 
quadratic in that new letter. 


‘Thus, in the equation, 

| | to/¢ =3. =D = 
which may be written 
| 2-3-vVz-3-2=0, 
if we let z = -/x — 3, wehave 2? —z-2 =0. 


Similarly, in 223 +4741 =0, 

if we let z = 2~*, we have 222 4+2+1=0. 

Again, in ax” + br” +c = 0, 

if we let z = x”, we have az? + bz +c =0. 
EXERCISES 


Solve the following equations and check results by substi- 
tution: 


1. 4—-38Vr%4+2=0. (1) 
SoLution : Let Vz = 2. 

Then GL = 2") 

Substituting in (1), 22 —32+2 +0. 

Factoring, (z — 1) — 2) =0. 

The roots are 2 .= 1, 2 = 2. 

Thus, Vze=1, vax =2, 

or oe 1,z =4. 


CHECK: 1 Sai 0;4-3-242 20. 


Art. 124] EXERCISES 205 


2,2." — t127-7 + 30 =.0. (1) 

SotuTIon : Let a =2. (2) 
Then from (1) and (2), 22 —11z + 30 =0. 
Factoring, (2 — 5) (z — 6) =0. 
Then Zim), Zim G, 
Hence, : =+¥V5, : =+V6. ~ (3) 
and + ya AIS 

V/5 5 V6 6 

The four numbers + ied + v6 ~ wy are solutions of (1). The 


phe eee 6. a G 
student should check by substitution in (1). 


3. z+ + 27 — 20 = 0. (1) 
SoLuTion : Let a? = 2Z, (2) 
Then from (1) and (2), 2274+2z—-—20 =0, (3) 
(2 —4)(2 +5) =0. (4) 

Then g-= 4,2 = —5. (5) 
Hence, g=+2,2 = t+iV5. (6) 


Note that two of the four roots are imaginary. The student should check 
by substitution in (1). 





fo 2 — 56 = 0. (1) 
SoLuTion: Let BM = 2; (2) 
Then from (1) and (2), 22 —z— 56 =0. 
Solving by the formula, z = ee = 8or -7. 
Hence, ; z= V8 or V—7. 


Curck : (W8)2" — (x/8)" — 56 = 8? — 8 — 56 =0. 
ogee (4/ — 7)" — 56 = (— 7)? +7 = 5650. 
5. 2* — 182? + 36 = 0. 8. 2 = Ore te = 0. 
6. 2? +3r71+2=0. 9. Vz +10+ Vx + 10 = 2. 
7 2—2zt =6. 10. 2+V72x+6= 14. 





206 


EA 


EQUATIONS IN QUADRATIC FORM ([Caap. XIV. 


= 15 
eh ar ae, 


t 


: (s+) +4047 = 12, 
x = 

. 23 — 972 +8=0. 

» (2? + 52)? + ale tb = 42: 


(4a? — 3)? + (8a? — 6)? = 80. 


. A/ 2a a ea 

. 3V2? 4+ 6VWe —4=0. 

. 2 + 3la-4 — 32 = 0. 

. tab Se +897 ob oo 4 ee 
92? = 4n8V/ a = 22-6 ie 
Ze 


A Dy eer fie ee 


Hint: Write the equation in the form 2! + 223 + 2? — 2(z? +2) 
—3 =0 and make z?+2 =z. 


22. 
23. 
24. 
25. 
Solve the following equations for log x: 
29. 
30. 


vt — 843 + 23x? — 28x — 8 = 0. 


One Tra 20 0): 26. 3(p* —2]) =p 
a = /a72? —8a* = 0. oT. K(k? 1) 
p* — 10p? — 11 = 0. 28. 1627 (1 — 2) = 3. 


(log x)? — 5logx+4=0. 
3(log x)? — 11 log x —4 = 0. 


Solve the following equations for 10*. 


31. 
32. 


10% — 7.107 +6 = 0. 
10% — 11.107 + 10 = 0. 


33. If the square of a certain number is multiplied by this 
square increased by 6, the result is 40. Find the number. Are 


there any integers that satisfy this condition ? 
numbers? Check results by substitution. 


Any imaginary 


Arts. 124, 125] EXERCISES 207 


34. If the square of a certain number is multiplied by this 
square decreased by 4, the product is 192. Find numbers that 
satisfy this condition, and check results by substitution. Are 
the results integral? Fractional? Real? Imaginary ? 


125. Simultaneous quadratic form. Any system of equa- 
tions is in simultaneous quadratic form with respect to ex- 
pressions containing the unknowns, if we may replace such 
expressions by new letters, and obtain a system of simultaneous ~ 
quadratics in such new letters. 

Thus, in the system, 
eg? —y =6, 
gt — yl =2, 


if weletz = 271, w = y~, we have 
22 — w* = 6, 


z2—w =2, 
EXERCISES 
Solve the following equations and check by substitution : 
Lae = yo? = 6, (1) 
gt=-yt = 2. 
SoLuTion : Let Po Bex. 2, Yt =a, 
Then system (1) becomes 22 — pw? = 6, (2) 
2—-w =2. (3) 
Solve for z in (3) and substitute in (1), 
w? +4w +4 -— w* = 6, (4) 
4w = 2, (5) 
; w =}. (6) 
From (6) and (3), z=3. 
Hence, 2=4,y =2. 
CHECK : (4)? — (2)? = (5)? — (a)? =6, 
(Q)?-Q)t=5 4 =2. 
1 1 1 1 
Beas, +> Ses 
1 Loin 
— = §, ne 3 


208 








4. = Bt 7 e£+y = 18, 
cogil Va-Vy =1. 
eee 
——-—=2, 
ais 8 2+ Vy =7 
x = 7, 
cis Oeste : Sal 
“Ot? un ty aed) Sia) Ma Vy —-Vai =1, 
1 1 
preemie 1 Mis VL 2 2 
as » (e349 
Git : 6 ; i 
er 229 uae 
ry = 1. ee pe 
EXERCISES AND PROBLEMS 
Solve the following equations for x: 
1. 424 — 52? +1 =.0, 2. 324 — 7x7 -6 = 0. 
3. vz — 13x? + 40 = 0. 
4, (x? — 1)? — 38(7? —1) +2 =0. 
3 x—5 
a se ye 
5.2 2 2. Lem x — 9d. 
pA: ae 8. 2 — 1laz? + 24 = 0. 
oxo ees 12: 9. 10zt — 4x? — 32 = 0. 
10. (32 + 4)? + 5(82 + 4)4 —6=0. 
11. x* — 4(a + b)a? + 4(a + OD)? = O. 
12. xt — 4(a + b)x? + 16(a + Bb)? = O. 
18.. 35257 -— 12a tel = 0: 
14. (a? — 5x + 6)? — 8(2? — 54 + 6) +: 2 = 0. 
15. (2? — 8)(a? — 3) = 374. 
16. (47? — 5¢ + 2)? = 122? — 1dr + 4. 
17. c— 15V/x + 56 = 0. 
as 
18. =r — 4q% er = 0. 
2 


EQUATIONS IN QUADRATIC FORM ([Cuap. XIV. 


Arr. 125] EXERCISES | 209 


Solve the following equations for x and y: 


i gl as ee O 
et 5 cee = 3° 
) ha 
gy? 4 23. (x — 2y)? — 2 + y = 6, 
82 — 5y = 11. 
Veg 
20. -+- = 4, x Yy 
my BA ee 8; 
fe 17, ants 
x? y2 2 ce agai § 
sro 
21.x2-y=2, . 25. x? + dy? = 7, 
Vi + Vy =2. ca a hh 


26. (x — 4)? — (y — 6)? = 0, 
(x —4)?+y =8. 


27. George Washington was born a.p. 1732. In the year 


A.D. «, he was x — 10 years old. Find the year a.p. zx. 

28. Find a number whose square multiplied by its square 
plus four gives 1440. 

29. Find two numbers the sum of whose reciprocals is 5, 
and the sum of the squares of whose reciprocals is 13. 

30. Find two numbers whose difference is 36, and the differ- 
ence of whose square roots is 2. 

31. The difference of the reciprocals of two numbers is 5, 
and the product of the numbers is .05. Find the numbers. 


CHAPTER XV 
PROGRESSIONS 


126. Arithmetical progressions. An arithmetical progres- 
sion 1s a succession of numbers each term of which, after the 
first, differs from the next preceding term by a fixed number 
called the common difference. 


Thus, 2 AAG AR 


is an arithmetical progression with the common difference 2. In the arith- 


metical progression 
10, 8, 6, 4, 2,-- > 


the common difference is —2. 

12%. Elements of an arithmetical progression. Let a repre- 
sent the first term, d the common difference, n the number of 
terms considered, / the nth, or last, term, and s the sum of the 
terms. The five numbers a, d,n,l, and s are called the elements 
of the arithmetical progression. 


128. Relations among the elements. Since a is the first 
term, we have, by definition of an arithmetical progression, 
a +d = second term, 
a + 2d = third term, 
a + 3d = fourth term, 


a+ (n — 1)d = nth term. 
That is, L=at+(n-A)d. (1) 
The sum of an arithmetical progression may be written in 
each of the following forms : 
a+(a+d)+(a+2d)+---+(l-2d) + (l-—d) +], 
1+(l—d)+(l—2d)+---+(a+42d)+(@+4+d) +a. 


S 


s 


‘ we have 8 =3 + 7d, ord = 


Arts. 128, 129] ARITHMETICAL MEANS 211 
By addition 


2 =(a@+)+(a+l)+(@t+)+--:: +(a4+)) 
=n(a+l). 
Therefore, sa Sa +1): (2) 


Whenever any three of the five elements are given, equations 
(1) and (2) make it possible to find the remaining two elements. 


129. Arithmetical means. The first and last terms of an 
arithmetical progression are called the extremes, while the re- 
maining terms are called the arithmetical means. ‘To insert 
a given number of arithmetical means between two numbers 
it is necessary only to determine d by the use of equation (1) 
and to write down the terms by the repeated addition of d. 


Example. Insert 6 arithmetical means between 3 and 8. 


SoLtuTIon: We have to find d, when a = 3,1 = 8, andn =6 42 =8. 
Since Ll =a + (n —1)d, 

5 

= 


Hence, the 6 arithmetical means between 3 and 8 are 


26 31 36 41 46 SL 
(EG NS ay IY SR | 


EXERCISES 


Find / and s for the following five exercises: 
1.2+11+20+--:-: to 10 terms. 


SOLUTION : Ll=a-+(n —1)d. 
Here, a =2,d =9,n = 10. 
~=2+9-9 =88. 
n 
s = 5 (a +1), 
= §(2 + 83) = 425 


ZLe PROGRESSIONS [CHap. XV. 
2. —-4-—8-—12-—--- to 19 terms. 
D se tied 
3: 3 Aegan ee a - ++ to 21 terms. | 
1 840e Fol 
4. = O51 Ae - to 17 terms. 
bed pals : 

5. riba ri} ost fit Be - + + to 25 terms. 

6. Given d= 3,7 = 25,1 = 74 findiamuae 

7. Givenl=—4,n=14,d =4; finda ands. 

8. Given a = 3, t = 108, s = 1221 > Tita eee 

9. Given a = 2,1=9,d =4; find n and s. 
10. Given a = —4, | = —64, n = 21; find d and s. 
1 74 1633 

11. Given d = 5) l= 7 Doreen y bs find a and n. 
12. Giver | = 41, n = 41;.8 = 8613 finda 
13. Given a = —2,d = o s = 0.; find mance 
14, Given a =7,n=7,8 =7; find d and 
15. Given d = 10, n = 10, s = 10; find qand?®. 
16. Insert 5 arithmetical means between 3 and 21. 
17. Insert 8 arithmetical means between 9 and 81. 
18. Find the arithmetical mean between 84 and 162. 
19. Insert 11 arithmetical means between 2 and 20. 


130. Geometrical progressions. A succession of terms in 
which the same quotient is obtained by dividing any term by 
the preceding term is called a geometrical progression. This 
quotient is called the ratio. 


Thus, 


Brame, 24, <.° 


is a geometrical progression with a ratio 2. 


| 


. 
4 


Arts. 131, 132, 133] GEOMETRICAL PROGRESSIONS 213 


131. Elements of a geometrical progression. The elements 
are the same as those for an arithmetical progression with one 
exception. Instead of the common difference of an arithmet- 
ical progression, we have here a ratio represented by r. 


132. Relations among the elements. If a represents the 


first term, then 
second term, 


Q 
— 
I 


ar? = third term, 


a 

be 
ds 
| 


= fourth term, 


ar’ = nth term. 


That is, if / represents the nth term, we have 


t=ar!, (1) 

By definition, 
s=a+t+art+ar+ar4+:-++ar, (2) 
Then, sr=artar+ar+-+++ar-t-+ar’. (3) 


Subtracting members of (2) from members of (3), we have 


sr —s=ar —a. 


Since 1 = ar”, (4) may be written in the form 


ri a. 
pa ess (5) 





Here, as in an arithmetical progression, whenever any three 
of the five elements are given, relations (1) and (5) make it 
possible to find the other two. 


133. Geometrical means. The first and last terms of a 
geometrical progression are called the extremes, while the 
remaining terms are called the geometrical means. To insert 
n geometrical means between two given numbers is to find a 


214 PROGRESSIONS [Cuap. XV. 


geometrical progression of n + 2 terms having the two given 
numbers for extremes. 


Example. Insert four geometrical means between 2 and 64. 


SoututTion: In this case 


a= 2, 

lL = 64, 
and n=. 
Then, by (1), Art. 132 64 = 27°, 

32 = 7", 
and | pes ps 


Hence, the four geometrical means are 4, 8, 16, 32. 


| EXERCISES 
1. Given a = 5,r =3,n=10; findland s. 
2. Given a = —3,r = 2,n =8; findl ands. 
8. Given a = 4,r = 4,n = 10; findl and s. 
4. Given.a = 5,71 = —2, ee find J and s. 
5. Givens — 50a = 670 = 10+ ind eee 


6. The third term of a geometrical progression is 3, and the 
sixth term is 81. What is the tenth term ? 

7. What is the fifth term of a geometrical progression whose 
first term is 2 and third term 4? 
8. What is the sum of the first five terms of a geometrical 
progression whose first term is 2 and third term 8 ? | 

9. The first term of a geometrical progression is 4, and the 
last term 256. If there are four terms in the geometrical pro- 
gression, find the common ratio and the sum of the series. 

10. Insert one geometrical mean between 6 and 150. 

11. Insert two geometrical means between 2 and 250. 

12. Insert three geometrical means between 12 and 2 

13. What is the eighth term of the series 3, $, yy,...? 

14. If each term of a geometrical progression is multiplied 
by the same number, show that the products form a geometrical 


progression. 


Art. 134] INFINITE NUMBER OF TERMS 215 


134. Number of terms infinite. Consider the geometrical 

progression 

Z) zy % ie boas 

It may at first thought appear that the sum of the first n terms 
could be made to exceed any finite number previously assigned 
by making n large enough. That this is not the case and that 
the sum can never exceed unity, will be seen from the following 
illustration. Conceive a particle moving in a straight line 
toward a point one unit distant in such a way as to describe 4 
the distance in the first second, $ the remaining distance in the 
second second, 4 the remaining distance in the third second, 
and so on indefinitely. This is represented in Fig. 52. 

The distance AB represents one unit of distance. In the 
first second the particle moves from A to P;. In the second 
second it moves from P; to Ps, and so on. The total distance 
traversed by the particle in n seconds is given by the sum 

,+4+4+-°::: ton terms, 
which sum cannot exceed nor equal 1, no matter how many 
terms we take, but can be made to differ from 1 by as small 
a positive number as we please by making the number of terms 





large enough. Thus, when n = 10, the sum is ae (Exercise 3, 


1024 
Art. 133). In this illustration, 1 is said to be the limiting value 


Ys P, P3; Pi Ps 
Se eT TT le eet 
A B 

Fig. 52 


of the sum of the first n terms of the progression. If s, repre- 
sents the sum of the first n terms, we write 

lim 

1 = 00 


amt 


which reads, “‘ the limit of s, as n increases indefinitely is 1.” 

The sum s of the infinite progression is defined as this limit. 
For any geometrical progression in which the ratio is less 

than 1, the above argument can be repeated, and it can be 


216 7 PROGRESSIONS [Cuap. XV. 


shown that there is a limiting value to the sum of the first n 
terms of such a progression. In Art. 132, we have shown that 
the sum, 

a+ ar + ar? +.--+*-- are, 


is given by Sn = a Sn) 


We then write 


lim » shim a. ligne eee 
hi 2 sn = 





We shall consider only the case where r < 1. In this case, 
as n increases, the value of r” becomes smaller and smaller. In 
arr 

—r 
value than any number that has been assigned and left fixed, or 


is smaller in absolute | 





fact, we may take n so large that i 
a nm 

l-r 

Hence, the sum of the infinite geometrical progression 





as we say, approaches the limit 0 as n becomes infinite. 





a, At, OF: <= , where 1. 
is s = lim s, = ° 
n=o 1—r 
EXERCISES 
Find the sum of the following infinite progressions : 
1. Ota we ar 
SoLuTION: Here ai=6, 7 =A. 
a 6 
desk. Shree | 
jess ten | Tel 
2. 1 OY 4 Rie 5. 27 30 
7 Rag be | 6. .5,'.125) 03125 
. Be Q’ 27” Pe 1 1 
4229-16, Radere 1. V2, 1 a oe 


8. 2, 25, 03125 0am 9. .3, .03,c00aaaee 


Arts. 135, 1836] HARMONICAL PROGRESSIONS 217 


135. Repeating decimals. Repeating decimals furnish good 
illustrations of infinite geometrical progressions. For example, 
.66666 - - - may be written as 


6 + .06 + .006 + .0006 + .00006 +: - - 


where a = .6 and r =.1. The limiting value of the sum of n 
terms as the number n increases indefinitely is 3. 
Again, .4919191 - - - may be written 


A+ .091 + .00091 +--- 


where the terms after the first form a geometrical progression 
in which a = .091 and r = .01. 


EXERCISES 


Find the limiting value of each of the following repeating 
decimals : 


eit >. 6. .83333 --. 11. .404040--. 
20800 - hy ea Pa Pee 12. .242424 --. 
oo 2LbG66 >. 8. .005050 - -. 13.. 1.181818 --. 
4. .363636 --. 9. 234234 --. 14. 2.272727 --. 
fereuulU*:. 202 OUSOU ee. 15. 3.363636 - -. 


136. Harmonical progressions. ‘Three or more numbers are 
said to form a harmonical progression if their reciprocals form 
an arithmetical progression. The term ‘ harmonical”’ as here 
used comes from a property of musical sounds. If a set of 
strings of uniform tension whose lengths are proportional to 
1, 4, 4, +, $, , - - - be sounded together, the effect is harmoni- 
ous to the ear. The succession of numbers 


i. se 
Vege). 5). 6.) 
is a harmonical progression since the reciprocals form the 
arithmetical progression 
Bee’ <. 
ly 2, 3, 4, v) 6, 


218 PROGRESSIONS [CHap. XV. 


13%. Harmonical means. To find n harmonical means 
between two numbers, find n arithmetical means (Art. 129) 
between the reciprocals of these numbers. The reciprocals of 
the arithmetical means are the harmonical means. 


EXERCISES 


Insert a harmonical mean between 4 and 3. 
Insert two harmonical means between 1 and 4. 
What is the harmonical mean between x and y? 
4. Show that 3, 2, 3 are in harmonical progression and 
continue the series two terms in each direction. 


SOF PSs 


MISCELLANEOUS EXERCISES AND PROBLEMS 


Find the seventh term of the progression 18, —6, 2, -- 

Find the fifth term of the progression a, a + d,a + 2d, -- 
1 

n+ n+2n + Tee 

What is the arithmetical mean ee. a and 6? 

Insert 5 arithmetical means between —4 and 20. 

Insert 3 geometrical means between 5 and 405. 

What is the geometrical mean between a and 6? 

8. The fourth term of a geometrical progression is 108, the 
sixth is 972. Find the ninth term. 

9. The sum of three numbers in arithmetical progression is 
15; if 1, 4, 19 are added to the numbers, the results are in 
geometrical progression. Find the numbers. 

10. Show that the sum of the first n odd numbers 1, 3, 5, 7, 

-is n?. 


Find the tenth term of the progression ~ : 





oN LLE  aa hd pete edt A ade pe 


PROBLEMS PERTAINING TO MOTION 


11. If a body falls 16 feet the first second, 48 the next, 80 
the next and so on, how far does it fall in the twelfth second ? 
How far has it fallen at the end of 12 seconds? 


Art. 137] PROBLEMS 219 


12. If a body falls 16 feet the first second, 48 the next, 80 
the next and so on, how far does it fall in the tth second? How 
far has it fallen at the end of the tth second ? 

13. A ball rolling down an incline of 30° goes 8 feet the first 
second, and in each second thereafter 16 feet more than in the 
preceding second. How far will it roll in 10 seconds? 

14. A marble rolls down an inclined plane, passing over 
distances 3 feet, 9 feet, 15 feet, in successive seconds. How 
long will it take it to pass over 108 feet ? 

15. Assume that a ball falls 16 feet the first second, 48 the 
next, 80 the next, and so on. A baseball was dropped from the 
top of Washington Monument, 550 feet high, and caught by 
an American League catcher. About how fast was the ball 
falling when caught ? 

16. In a potato race there are placed 20 potatoes, at the 
distances 5 feet, 8 feet, 11 feet, and so on, from a bag. A man 
starting from the bag is required to pick up the potatoes and 
carry them back to the bag one by one. What is the total dis- 
tance that he goes in thus collecting the potatoes ? 

17. A wheel of perimeter 6 feet, free to rotate on an axis, 
is started to rotate 20 revolutions per second, and then is to be 
retarded by friction. If it makes 95% as many revolutions 
each second thereafter as it did the preceding second, how far 
will a point on the rim have moved when the wheel comes to a 
standstill ? 


Hint: These distances are assumed to form an infinite progression. 


18. If a particle moves in a straight line from a given posi- 
tion, with such a speed that during any given second it moves 
60% as far as it did during the preceding second, and if it 
moved 25 feet during the first second, what is the limit of 
the distance the particle will move? 

19. If a ball should fall 9 feet and bound back 6 feet, then 
fall 6 feet and bound back 4 feet and so on indefinitely, what is 
the limit of the distance through which the ball would move? 


220 PROGRESSIONS [Cuap. XV. 


20. A travels uniformly 25 miles a day; B starts at the same — 
time, and travels 5 miles the first day, 10 miles the second, 15 
miles the third, and soon. In how many days will B over- 
take A ? | 


PROBLEMS PERTAINING TO BUSINESS 


21. Find the present value of twenty annual payments of 
$1000 made at end of each of the next twenty years, if money 
is worth 5% compounded annually. 


$1000 


Hint: The present value of $1000 due in n years is (1.05)"" 


22. Find the present value of $125 per year paid at the end 
of each year for forty years if interest is at 4% compounded 
annually. 

23. I owe a debt of $4675, to be paid in instalments, 
the first payment to be $800, the second $725, and decreasing 
by a common difference, until the last payment which is $50. 
Find the number of instalments. 

24. A gentleman being importuned to sell a fine horse, said 
he would sell him on the condition of receiving 1 cent for the 
first nail in four shoes, 2 cents for the second, and so on, doub- 
ling the price of every nail; the number of nails in each shoe 
being 8, how much would he receive for his horse ? 


CHAPTER XVI 
THE BINOMIAL THEOREM 


138. Factorial. The symbols 


r! and |r 
are read “ factorial 7,”’ and are used to indicate the product 
1-2-3-...r. The symbol r! is easier to set up in type 


than the symbol |r, and is used more in print than |r. 
Thus, 3! =1-2-3=6; 7!=1-2-3-4-5-6-7 = 5040. 


139. Powers of a binomial. By actual multiplication, 


(a + b)? = a? + 2ab + B?, (1) 
(a + b)? = a3 + 3a%b + 3ab? + B3, (2) 
(a + b)* = at + 4a%b + 6a7b? + 4ab3 + Dt. (3) 


Exercise. Expand (a + b)* by actual multiplication of (3) by a + b. 


140. The expansion of (a +b)”. The expressions (1), (2), 
and (3) of Art. 139 are special cases of the general formula 


PA sees | 
(a+b)" =a" +na"" b+ Bee a” 2 


if n(n — 1) (n —2) 


31 a"—3$3 as oe, (4) 





This is called the binomial formula or binomial theorem. 

The following properties of this formula should be noted: 

(1) The first term is a”. 

(2) The second term is a”b. 

(3) The exponents of a decrease by unity from term to term 
while the exponents of b increase by unity. 


222 THE BINOMIAL THEOREM [Cuap. XVI. 


(4) If in any term the coefficient be multiplied by the ex- 
ponent of a and divided by the exponent of 6 increased by 1, 
the result is the coefficient of the next term. 

(5) The factorial number in the denominator of the coeffi- 
cient of any term is equal to the exponent of b in that term. 

(6) The numerator of the coefficient of any term is the prod- 
uct of factors beginning with n, decreasing successively by 1, 
and ending with n — r + 1, where r is the exponent of Db. 

(7) When n is any positive integer, the expression contains 
n + 1 terms of which the last is 6”. 

The binomial theorem holds for any positive integral value 
of n, and, under certain restrictions on a and b, for negative 
and fractional values of n. (See Art. 142.) 


The theorem is proved in more advanced algebra, but here it 


is merely stated without proof. 


EXERCISES 
1. Expand (a — 6)”. 


SoLuTion: Put —b for b in (4), Art. 140. The even powers of —b are 
positive and the odd powers are negative. We may therefore write 


(n — 1) 


(a — 6)" =a" — na™b + a oy a"—2b2 


= ne asb) 45s oa 


The last term is either —b" or +b" depending upon whether v is odd or 
even. 
2. Expand (8y + 52)5. 
SoLuTION : Substitute 3y for a, 5% for 6, and 5 for n in (4), Art. 140. 





| 4 cis eae 

(By + 52)> = (By)® + 5@y)! 5x +2~5 (8y)* (5x)? + 2-2-8 (By)? Gn) 
5-4-3-2 | 

+ [Sey Gy) (6x)! + (Gx)® = 243y5 + 2025 yx + 6750y%x* + 11250y%x? 


+ 9375ya* + 312525. 


Art. 140] EXERCISES 223 


6 
3. Expand (5 _ = 


x 


SoLuTION: Substitute ; for a, 2 for b, and 6 for n in (4) Art. 140. 


Then 
ges 2p \s e\* cy 2b EN? 2b\? c\3 2b\3 
pee) = (a) +6(5) (- 7) +¥5(5) (- 2) +20(5) (- 2) 
¢\?/ 2b\4 c 2b\° 2b\e 8 bc 
+15 (5) (-2) +6(5)(-2) +(-2] E7200 Sie 


20b%c! 160b%c?_ 80b'c? = 64b°c =~ 640° 











2722 273 <i Sat ist x6 
Expand : 
4, (x = ae. 15. (re ~+/x)4. 
5B. (1 + 2-7)”. 6 
\ ) 16. (5 — bx) . 
6. (a? — 2y)®. 2 
__ pnl)\5 1 way 
Roni — et)". i. (+ v2) 
8. (Qe ++/x)4. 
6 
1 a) 18. (2 “- *) 
: cc 2 
at 19. (a! + y')8§, 
aoe ais “) 20. («+ y +2)%. 
Q\7 Hint: Consider x + y as represent- 
11. (1 + =.) ing one term of a binomial. 
Tea t ; 
12h /2.—/3)°- 21. (5 > ing 1) 
tan? 
afte Pe ine 
13. (v2 - =) oR *) 
eA) 22 \ve + 4 +4 e 


14. (2 + 2vV/x)4. 23. (x! + 2y*)4. 


224 THE BINOMIAL THEOREM [Cuap. XVI. 


Use the binomial formula to compute the following, correct 
to three significant figures : 


DAS Lok) e 

Hint: Write (1.1) = (1 +.1)% and expand to a few terms. 
25. (.99). | 

Hint: Write (.99)2? = (1 — .01)”° and expand. 

26. Se) ae 27. (.999)!° 


141. Formula for the 7th term of the expansion of (a + b)”. 
We have seen by Art. 140 that the third term is 








n(n —1) 
eye 2b? 
the fourth term is eee ar h*; 
and hence that the rth term is 
n(n —1)(n—2) ... m—7r+4+2) toon 
(r — 1)! 
EXERCISES 
y\10 
1. Find the sixth term of (21 — y) . 
- Sotution: Here a = 22, b = —3y, n = 10, r = 6. 
x _10-9:8-7:6 .\.f y\5 __ 896 
The sixth term is [2a g.4 05 (22) ( u) =e ee 
2. In (a — y)", find the sixth term. 
8. In (a? — y?)™, find the middle term. 
4. In (x? — y*), find the middle term. 
5. In (3a — 4a?)!5, find the sixth term. 


5S ai¢ : 
6. In (2 i *) » find the middle term. 


7. In (.5\/a — .1v/y)", find the fifth term. 


Arr. 142] EXERCISES | | 225 

142. Binomial formula, any exponent. The binomial for- 
mula (a Mie b)” = qr + nab ae ae a”2h2 a8 

n(n 2% 1) (n a 2) qr 3h3 a 
3! 
holds when n is a negative or fractional number, provided the 
number 6 lies between —a and a. 

If n is not a positive integer, the formula for the development 
of (a + 6)” contains an unlimited number of terms. If b is 
small compared to a, the first few terms of the expansion give 
a useful approximation for certain purposes. Thus, the first 
few terms in the binomial expansion of (27 + 1)} gives the ap- 
proximate value of the cube root of 28. (See Example 3 below). 


EXERCISES 
1. Expand to four terms (a + y). 


SOLUTION : Substitute in the expansion for (a + 6)”. This gives 


ug 1) Lilia Vi dee 
cess 1 4-1 ma ag oe ne a 3- 
1% 1 oa 8 
ae, ae y- 5 $2 + 352 ys 


2. Expand to three terms (32 — 4y)~”. 
So.tuTion: By the formula for (a + b)", 


(3e - 5) . = (37)~? — 2(82)- (5 v) ee pe (@x)*{ - su) t+ 


il y 


ly “2 wad 
Oy? Ai 2723 T 082! 





+. 


3. Find approximately 7/28 by the binomial formula. 
SoLuTION : Write 


35-1) 
0/28 = (27 +1)? = 278 +3 F o7-8 4 3(3 ~ See O74 


Bee ical 
t 97 — 9187 7 
= 3.0366 —. 


226 THE BINOMIAL THEOREM [Cuap. XVI. 


Expand the following expressions to four terms: 


4, (1 — x)=. 10. (a?y-? + 1). 

Ba — 2) 11. (8a* + 1). 

6..(2a;+ 1)>. 12. (844 = 

poner i es 13. (9x? + 27) 
Al 

8 (1 — 2). 14, (« _ *) ; 

QA eine 15. 7/9 = (84 1)% 


Find the roots indicated, correct to three significant figures : 
16. +/10. 18. (79)}. 20. /730. 
17. (26)3. 19. 1/65. 21. (25)}. 


Art. 142] REVIEW EXERCISES AND PROBLEMS 227 


REVIEW EXERCISES AND PROBLEMS ON CHAPTERS XIII-XVI 


1. Find the arithmetic mean, the geometric mean, and the harmonic 
mean between a and b. 

2. In a Christmas savings club there are three classes of depositors. 
One class deposits 1 cent the first week, 2 cents the second week, 3 cents 
the third week, and so on, increasing the weekly deposit 1 cent each week 
for 50 weeks ; the second class deposits 2 cents the first week and increases 
the weekly deposit 2 cents each week for 50 weeks ; the third class de- 
- posits 5 cents the first week, and increases the weekly deposit 5 cents each 
week for 50 weeks. Find how much is deposited by a member of each 
class in 50 weeks. 

3. The population of a town is P at a certain time. Annually it loses 
2% by deaths and gains y% by births. What is the population at the end 
of n years? 

4. What is the outer diameter of a spherical shell that is 1 inch thick 
and contains 50 cubic inches ? 


5. Solve for x (a) x? + 2a? = daz. 


(b) a? +327 = 4. (PRINCETON) 


uff 2n : 
6. What term in the developmént of (a =) does not contain a. 


(SHEFFIELD) 


7. Leta, b, c (Fig. 53) be three verticals erected so that a line from the 
foot of a to the top of c, or from the foot of 
c to the top of a, bisects the intermediate 
vertical b. Show that b isthe harmonic mean 
between a and c; that is, show that 


G43) 3 





Fic. 53 8. In Fig 54 the lines a, b, c, d,+-- 
are drawn so that the angles marked 1, 2, 
3, ° ++ are equal. Show that the lengths of a, b, c, d,--- form a geo- 


metrical progression. 
Hint: Show that ; = ° by use of similar triangles, p. 99. 


9. Solve (a +11)'+ a? = 552-3. (ILLINOIS) 


10. Solve Ve+-1+2 =11. 


228 THE BINOMIAL THEOREM [Cuar. XVI. 








Fiac. 54 


11. Find to two places decimal by the binomial theorem the value of 
(120). (CALIFORNIA) 
12. A piece of land is in the form of a trapezoid, with parallel sides 
20 rods and 30 rods, and altitude 20 rods. It is 
desired to divide this piece of land into two 
equal parts by a line parallel to the parallel 
sides. Find the distance between the shorter 
parallel side and this dividing line. 


13. Find the value of log;49 + 4log, 64 + 
logs 216. Fig. 55 

14. Insert two arithmetical means, two geometrical means, and two 
harmonical means between 3 and 24. 

15. Find the fifth term of (2 +=3) x and reduce to the simplest 
form. (DARTMOUTH) 

16. Solve for z, (log x)? — 11 log x + 10 =0. 

17. The following quaint problem was found in an old Hindu manuscript: 
The square root of half the number of bees in a swarm have flown out upon 
a jessamine bush ; $ of the whole swarm have remained behind ; one female . 
bee flies about a male that is buzzing within a lotus flower into which he 
was allured in the night by its sweet odor, but is now imprisoned in it. 
Tell me the number of bees in the swarm. 

18. Solve for x and y 

(x +2)? — (2y - 3)} = 2, 
x—2y+3 = 14. 


Art. 142] REVIEW EXERCISES AND PROBLEMS 229 


19. Find the sum of n terms of the series, 


ed i Sen He 
+ + 
n n 








1+ 





-f- ee 
(Mass. INSTITUTE) 

20. A rectangle has the base x and altitude y, and a constant area 12. 
Construct the rectangles, on coédrdinate paper when x has the values 24, 
12, 8, 6, 4, 3, 2, 1, .5, .4, .3. In each case take’ the origin as ohe vertex 
and the codrdinate axes as sides. After all the rectangles are constructed 
draw a curve through the vertices that are opposite the origin. 

21. Gerbert (1003 a.p.), who was later: Pope Sylvester, found the alti- 
tude of an equilateral triangle by multiplying the side by # ; Heron (about 
100 B.c.) used the factor +3 ; the factor 7 is now used to make a rough ap- 
proximation. Which value is nearest the correct one? 

22. Continue the progression 3, 1, —4 two more terms and find the sum 


of the first 13 terms. | (ILLINOIS) 
23. Find the value to four decimal places of ae by the binomial 
theorem. (Mass. INstrTuTE) 
24. Insert an arithmetical mean between 97 and 47, also a geo- 
metrical mean. How do they compare ? (DARTMOUTH) 
25. Solve for x 20 (log x)? — 23 log « +6 = 0. 
26. Solve for x 6(log x)? +17 log x +7 =0. 


27. A number is 12 greater than 4} times its square root. Compute 
its value to two places of decimals. Verify the smaller value. 
(COLLEGE ENTRANCE EXAMINATION BOARD) 
28. (a) Of a geometric progression, the sum of the first and second 
terms is 16, the sum of the third and fourth terms is 36. Find the ratio, 
and the sum of the first 6 terms. (b) Write 
the seventh term of the expansion of 
(1 + x)® by the binomial theorem. 


(CoLLEGE ENTRANCE EXAMINATION 
Boarp) 
29. Find the roots of 492+ — 70x? + 12 
Fig. 56 = 0 to three decimal places. 

30. The sum of an infinite geometric 
series is 4, and the first term is 6. Find 
the ratio and the sum of 4 terms. (YALE) 

31. In Fig. 56, the vertical lines are 
equally spaced. Show that their lengths 
form an arithmetical progression. 

32. In Fig. 57, the vertical lines are so 
spaced, that the lines drawn from the foot of 





230 THE BINOMIAL THEOREM [Cuap. XVI. 


each to the top of the next are all parallel. Show that the lengths of the 
verticals form a geometrical progression. 

33. A strip of carpet one-half inch thick and 29% feet long is rolled on 
a roller 4 inches in diameter. Find how many turns there will be, remem- 
bering that each turn increases the diameter by one inch, and taking as 


the length of the circumference */ of the diameter. (HARVARD) 
34. Solve Vx — Vy =2, (Vx —- Vy)Vxy = 30. (YALE) 
35. Solver —y —-Va —y =2,23 —y' = 2044. (YALE) 


36. Find the middle term of the expansion of (x — y)®. Express this 





Ue bt 4 2 
10 15 
term in its simplest form when z = es , and y = ve _* (HARVARD) 
3 5 


37. kixtract the square root of 38 — 12/10. (REGENTS) 


38. Solve Vora +1 =384+ V2 +2. 
39. Jind the sum to infinity of the progression 


2 1 
—8, - 5) BOS (Mass. INsTITUTE) 
40. Solve for x 102% — 107 —2 =0. 
41. Find the value of the repeating decimal .3727272 * - - in fractional 


form. 
42. Solve 3(x + 1)2 +. 23 + 322 + 32 +3 =0. 


INDEX 


[The numbers refer to pages.] 


Abscissa, 66 Cube root, 
Addition, 2 of numbers in Arabic figures, 
associative law of, 2 176 
commutative law of, 2 of polynomials, 174 
of fractions, 44 
Algebraic operations, 12 Degree, 
Argand, 116 of an expression, 25 
Arithmetical means, 211 of a term, 25 
Arithmetical progression, 210 Determinants, 80 
elements of an, 210 expansion of, 83 
Associative law, historical note on, 85 
of addition, 2 of the second order, 80 
of multiplication, 5 of the third order, 82 
Axes, solution of equations by, 80, 85 © 
codrdinate, 66 Diophantus, 93 
Discriminant, 132 
Binomial theorem, 221 Distributive law, 
Briggs, 198 of multiplication, 6 
Division, 8 
Cardan, 116 of fractions, 46 
Characteristic, 180 _ of monomials, 8 
determination of, 184 of polynomials, 9 
Commutative law, zero in, 9 
of addition, 2 
of multiplication 5 Elimination, 78 
Complex numbers 114 by addition and subtraction, 78 
conjugate, 114 by substitution, 79 
graph of, 117 Equalities, 15 
operations with, 114 Equations, 15 
Constant, 59 as sentences, 16 
Coérdinate axes, 66 equivalent, 16, 77 
Coérdinates, 66 graphs of linear, 74 


Cramer, 85 incompatible, 77 


232 


Equations (continued) 
inconsistent, 77 
indeterminate, 91 
in quadratic form, 204 
in simultaneous quadratic form, 
207 
involving fractions, 50 
involving radicals, 200 
linear in one unknown, 18 
operations on, 17 
quadratic, 1238, 128 
solution by factoring, 38, 124 
solved by determinants, 80, 85 
systems of, involving quad- 
ratics, 141 
roots of, 16 
with given roots, 130 
Evaluation of functions, 61 
Exponents, 6, 152 
fractional, 154, 155 
historical note on, 155 
laws of, 152 
negative, 155 
positive integral, 152 
zero, 154 
Extremes, 
of a proportion, 95 


Factor, 25 
highest common, 38 
Factorial, 221 
Factoring, 25 
solution of equations by, 38, 124 
summary of, 34 
Factor theorem, 30 
Formulas, 
frequently used, 62 
Fractions, 41, 108 
addition and subtraction of, 44 
algebraic, 41 
clearing of, 50 
complex, 48 


INDEX 


equations involving, 50 
multiplication and division of, 
46 

reduction of, 41, 42 
Function, 59 

evaluation of, 61 

_ graph of a, 68 

Functional notation, 60 
Fundamental laws, 

note on, 13 
Fundamental operations, 1 


Gauss, 116 
Geometrical means, 213 
Geometrical progression, 212 
elements of, 213 
infinite, 215 
Girard, 116 
Graph, 
of a complex number, 117 
of a function, 68 
of a linear equation, 74 
of logax, 185 ; 
of a pure imaginary number, 
116 
of a quadratic function, 134 
Graphical solution, 
of a system involving quad- 
ratics, 142 ; 
of a system of linear equations, 
76 


Hamilton, Sir William Rowan, 14 
Harmonical means, 218 
Harmonical progression, 217 
Highest common factor, 38 
Historical note, 
on determinants, 85 
on exponents, 155 
on fundamental laws, 13 
on imaginary numbers, 116 
on indeterminate equations, 92 


INDEX 


Historical note (continued) 
on logarithms, 193 
Hyperbola, 143 


Identities, 15 
as sentences, 16 
Imaginary numbers, 111 
graph of, 116 
historical note on, 116 
products of, 118 
pure, 114 
Indeterminate equations, 91 
historical note on, 92 
Integers, 108 
Integral expressions, 25 
Irrational numbers, 110, 159 


Leibnitz, 85 
Linear equations, 
graphical solution of a system 
of, 76 
graph of, in two unknowns, 
74 
in one unknown, 18 | 
solved by determinants, 80, 85 
Logarithm, 179 | 
characteristic of a, 180 
common or Briggs, 183 
computation by, 190 
historical note on, 193 
mantissa of a, 180, 185 
of a power, 181 
of a product, 181 
of a quotient, 181 
table, 186, 187 


Mantissa, 180, 185 
Mean proportional, 96 
Means, 
of a proportion, 95 
Members, 
of an equality, 15 


233 


Monomials, 

division of, 8 

division of polynomial by, 9 

multiplication of, 6 
Multiple, 

lowest common, 40 
Multiplication, 

associative and commutative 

laws of, 5 

by zero, 7 

distributive law of, 6 

of fractions, 46 

of monomials, 6 

of polynomials, 6, 7 


Napier, Baron John, 198 
Number concept, 
extension of, 108 
Numbers, 
complex, 114 
conjugate complex, 114 
imaginary, 111 
integers and fractions, 108 
irrational, 110, 159 
negative, 109 
rational, 110, 159 
real, 111 


Operations, 
algebraic, 12 
historical note on fundamental, 
13 
on equations, 17 
Ordinate, 66 
Oresme, Nicole, 155 
Origin of codrdinates, 66 


Parentheses, 
use of, 3 

Polynomials, 
division of, 9 
multiplication of, 6, 7 
square roots of, 172 


234 


Powers, 11 
of a binomial, 221 
of 7, 113 
Prime, 
expressions, 25 
expressions prime to each other, 
39 
Products, 
important special, 26, 29 
of imaginaries, 113 
Progressions, 
arithmetical, 210 
geometrical, 212 
harmonical, 217 
Proportion, 95 
by alternation, and by inver- 
sion, 97 
by composition, by division, 


and by composition and 
division, 98 
Proportional, 


fourth, mean, third, 96 


Quadratic equation, 123 
“nature of roots of, 131 

simultaneous, 141 
solution by factoring, 124 
solution by formula, 125 
special or incomplete, 128 
typical form of, 123 

Quadratic function, 123 
graph of, 134 

Quadratic surd, 161 


Radicals, 160 
addition and subtraction of, 
164 . 
division of, 169 
equations involving, 200 
index of, 161 
multiplication of, 166 
order of, 161 


INDEX 


reduction to same order, 166 
similar, 164 
simplest form, 163 
simplification of, 162 
Radicand, 160 
Ratio, 94 
Rational integral expression, 25 
degree of, 25 
Rational numbers, 110, 159 
Real numbers, 111 
Reciprocal, 42 
Repeating decimals, 217 
Roots, 11 
extraneous, 17 
index of, 161 
of equations, 16 
order of, 161 


Simultaneous equations 
(See systems of equations.) 
Solution, 16 
graphical, 76, 142 
Square root, 
of numbers in Arabic symbols, 
173 
of polynomials, 172 
of radical expressions, 173 
Stevin, Simon, 155 
Subtraction, 2 
definition, 3 
of fractions, 44 
Surd, 161 
quadratic, 161° 
Systems of equations, 
both equations of 
ax? + by? ++c=0, 145 
both equations quadratic, 145 
elimination in, 78, 79 
graphical solution of, 76 
incompatible, inconsistent, de- 
pendent, equivalent, 77 
involving quadraties, 141 


form 


INDEX 


Systems of equations (continued) 
one linear and one quadratic, 
141 


solved by determinants, 80, 85 


Transposition, 17 


Variable, 59 

Variation, 102. 
combined, 103 
constant of, 102 


inverse, 103 

joint, 103 
Verification, 

by substitution, 18 


Wallace, John, 156 
Wessel, 116 


Zero, 7 
as an exponent, 154 
in division, 9 


235 
































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